find peremiter of pqr

Step-by-step explanation:

50cm/40m = 0.5m/40m = 0.0125 = 1/80

Answer:

y = \(\frac{3}{4}\) (x - 2)² - 5

Step-by-step explanation:

the equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k ) are the coordinates of the vertex and a is a multiplier

here (h, k ) = (2, - 5 ) , then

y = a(x - 2)² - 5

to find a substitute (- 2, 7 ) into the equation

7 = a(- 2 - 2)² - 5 ( add 5 to both sides )

12 = a(- 4)² = 16a ( divide both sides by 16 )

\(\frac{12}{16}\) = a , that is a = \(\frac{3}{4}\)

y = \(\frac{3}{4}\) (x - 2)² - 5 ← equation in vertex form

Answer:

\(y=\dfrac{3}{4}(x-2)^2-5\)

Step-by-step explanation:

\(\boxed{\begin{minipage}{5.6 cm}\underline{Vertex form of a quadratic equation}\\\\$y=a(x-h)^2+k$\\\\where:\\ \phantom{ww}$\bullet$ $(h,k)$ is the vertex. \\ \phantom{ww}$\bullet$ $a$ is some constant.\\\end{minipage}}\)

Given:

Substitute the given vertex and point into the formula and solve for a:

\(\implies 7=a(-2-2)^2+(-5)\)

\(\implies 7=a(-4)^2-5\)

\(\implies 7=16a-5\)

\(\implies 12=16a\)

\(\implies a=\dfrac{12}{16}\)

\(\implies a=\dfrac{3}{4}\)

Substitute the vertex and the found value of a into the formula to create an equation in vertex form for the given parameters:

\(y=\dfrac{3}{4}(x-2)^2-5\)

A straight line can be written in the form of a two variable linear equation.

The equation of line passing through point (4,0) is  y = -1 / 17 x + 4.

To represent a straight line on a graph consider two points namely x and y intercepts of the line. To find x-intercept put y = 0 and for y-intercept put x = 0. Then draw a line passing through these two points.

Given that,

The line is passing through the point (4,0).

It is parallel to the line passing through (7, -3) and (-10. -2).

The slope of the line passing through (7, -3) and (-10. -2) is

m₁ = (-2-(-3)) / (-10 - 7)

m₁ = - 1 / 17

Since the slope of two lines parallel to each other is equal, the slope of the line passing through point (4,0) is - 1 / 17.

Thus, the equation of line passing through point (4,0) can be written as,

(y - 4) / (x - 0) =  - 1 / 17

=> y = -1 / 17 x + 4

Hence, the equation of the given line is y = -1 / 17 x + 4.

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answer:

if you multiply the 2 sides of the ratio by a number it will give you an equivalent ratio

5:8 = 10:16 multiplying by 2

5:8 = 15:24 multiplying by 3

5:8 = 20:32 multiplying by 4

5:8 = 25:40 multiplying by 5

5:8 = 30:48 multiplying by 6

5:8 = 35:56 multiplying by 7

and so on ...

Nothing that we can say regarding an we are unsure of whether the terms of an are rising or falling, a passes the Comparison Test and sequence, Since bn, an also converges.

A collection of numbers, or "terms," is referred to as a sequence. Examples of terms are 2, 5, and 8. Some sequences can be made endlessly lengthy by taking advantage of a particular pattern they follow. Use the example of 2,5,8 and add 3 to continue the sequence.

We are unable to comment on an.

This is due to the fact that bn converges, causing a to exceed bn, and as a result, we are unable to determine whether the terms of an are increasing or decreasing.

What can you say about an if a bn for all n? a passes the Comparison Test and converges.

This is due to the fact that since bn converges and a converges since bn, we may infer that the terms of an are likewise decreasing and that an also sequence since bn based on the comparison test.

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Answer:

47 degrees

Step-by-step explanation:

This is a parallel line, and this sort of angle is known as an alternate angle, which is a z shaped angle (can be a Z either way). Alternate angles are equal, therefore x=47 degrees too. Hope this helps :)

the first institution has 441 and the second institution has 189.

We can use algebra to solve the problem. We can start by assigning variables to represent the two institutions:

Let x be the multiplier of the ratio.

Then, we have:

7x is the first institution.

3x is the second institution.

From the problem, we know that the sum of the two institutions is 630:

7x + 3x = 630

Simplifying the left side, we get:

10x = 630

Dividing both sides by 10, we get:

x = 63

Now that we know the value of x, we can find the number of each institution:

First institution = 7x = 7(63) = 441

Second institution = 3x = 3(63) = 189

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If (X,T) is a second countable and T1.5 space, then (X,T) is a Hausdorff space. If X is a countable set and (X,T) is first countable, it may or may not be pathwise connected, zero-dimensional, or second countable.

10) If (X,T) is second countable and T1.5 space, then (X,T) is a Hausdorff space (option A). A topological space is said to be Hausdorff if for any two distinct points x and y in X, there exist disjoint open sets U and V containing x and y, respectively. The fact that (X,T) is second countable implies that there exists a countable basis for the topology T. Being T1.5 means that for any two distinct points x and y, there exist open sets containing x and y, respectively, such that their closures do not intersect. Since (X,T) satisfies both properties, it is Hausdorff.

11) If X is a countable set and (X,T) is first countable, it cannot be concluded whether (X,T) is pathwise connected (option A), zero-dimensional (option B), or second countable (option C). First countability means that each point in X has a countable neighborhood basis.

A countable set can have multiple disconnected components, so it may or may not be pathwise connected. Similarly, a countable set can have different topologies, and being first countable does not determine whether it is zero-dimensional or second countable. Therefore, the correct answer is option D, N.A. (Not Applicable) since the properties mentioned are not determined solely by X being countable and (X,T) being first countable.

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Answer:

See below.

Step-by-step explanation:

Since the two triangles are similar, this means that their corresponding sides are proportional. In other words:

\(\frac{PQ}{XY}=\frac{QR}{YZ}=\frac{RP}{ZX}\)

Part A:

We already know PQ, RP, XY, and YZ.

To find XZ, we can use the first and third proportions:

\(\frac{PQ}{XY} =\frac{RP}{ZX} \\=\frac{7.2}{18}=\frac{6.5}{ZX}\)

Cross multiply:

\(117=7.2(ZX)\\ZX \text{ or } XZ = 16.25\)

Part B:

The same thing as Part A, but with the first and second proportions:

\(\frac{PQ}{XY}=\frac{QR}{YZ}\\\frac{7.2}{18}=\frac{QR}{7}\\50.4=18(QR)\\QR=2.8\)

when running a line, in a right-triangle, from the 90° angle perpendicular to its opposite side, we will end up with three similar triangles, one Small, one Medium and a containing Large one, from where we can use proportions to get their sides, so Check the picture below.

Answer: x= 8\(\sqrt{5}\)

Step-by-step explanation:

hope it helps

Answer:

C yes, one to one

Step-by-step explanation:

for every different value of age there is one specific function result.

f(age) = summer camp group, age in N, 6 <= age <= 8

this function is therefore not defined for ages outside of that interval, or if not natural numbers.

under these criteria every allowed value of age has a different calculated summer camp group. no two values of age have the same result as summer camp group.

therefore, it is a one-to-one function.

Answer:

Step-by-step explanation:

To graph the equation 3y = -6x - 12, we can start by finding the x and y intercepts.

The x-intercept is the point where y = 0, so we can substitute 0 for y in the equation:

3y = -6x - 12

3 * 0 = -6x - 12

0 = -6x - 12

12 = -6x

x = -2

So the x-intercept is (-2, 0).

The y-intercept is the point where x = 0, so we can substitute 0 for x in the equation:

3y = -6x - 12

3y = -6 * 0 - 12

3y = -12

y = -4

So the y-intercept is (0, -4).

Now that we have the x and y intercepts, we can plot the points on the graph and draw a line that passes through both points. This line will be the graph of the equation.

In this case, the graph of the equation 3y = -6x - 12 is a downward sloping line that passes through the points (-2, 0) and (0, -4).

Take into account that the confidence interval is given by:

where

μ: mean = 18.2

Z: z-value for 95% confidence = 1.96

s = 18.2

n: sample = 86

Replace the previous values into the confidence interval formula:

Hence, the confidence interval is:

CI = (18.2 - 3.85 , 18.2 + 3.85) = (14.35 , 22.05)

Rounded to the nearest whole number the confidence interval is:

CI = (14 , 22)

Answer:

Approximately \(12.9\; \rm km \cdot h^{-1}\), assuming that this orbit is circular.

Step-by-step explanation:

The question is asking for a tangential velocity with the unit \(\rm km \cdot h^{-1}\). The unit of the given distance is already in \(\rm km\) as required. Convert the unit of the orbital period to hours:

\(\begin{aligned}T &= 1230\; \text{day} \times \frac{24\; \rm h}{1\; \text{day}}. = 29520\; \rm h\end{aligned}\).

Calculate the angular velocity \(\omega\) of this planet from its orbital period:

\(\begin{aligned}\omega &= \frac{2\, \pi}{T} \\ &= \frac{2\, \pi}{29520\; \rm h} \approx 2.1285 \times 10^{-4}\; \rm h^{-1}\end{aligned}\).

Given the radius \(r\) of the orbit of this planet, the tangential velocity \(v_{\perp}\) of this planet would be:

\(\begin{aligned}v_{\perp} &= \omega\, r \\ &\approx 2.1285 \times 10^{-4}\; \rm h^{-1} \times 2.22\times 10^{7}\; \rm km \\ &\approx 4.73 \times 10^{3}\; \rm km \cdot h^{-1}\end{aligned}\).

If the orbit of this planet is circular, the velocity of the planet would be equal to its tangential velocity: \(4.73\times 10^{3}\; \rm km \cdot h^{-1}\).

Answer:

385 miles

Step-by-step explanation:

Answer:

1

Step-by-step explanation:

because its the only one where if you run a line down the y axis it only crosses the line at one point

Answer:

The coordinates of the vertices of the image are (0, 0), (2, 3), (5, 0) ⇒ (A)

Step-by-step explanation:

Let us talk about dilation

→ If point (x, y) dilated with scale factor m about the origin, then

→ Its image is (m x, m y)

Let us solve the question

∵ ΔXYZ has vertices:

→ X = (0, 0)

→ Y = (6, 9)

→ Z = (15, 0)

∵ ΔXYZ is dilated by a scale factor 1/3 about the origin

→ Multiply each coordinate in each vertex by the scale factor 1/3

∵  (0 × \(\frac{1}{3}\) , 0 ×  \(\frac{1}{3}\) )

∴ X' = (0, 0)

∵  (6 × \(\frac{1}{3}\) , 9 ×  \(\frac{1}{3}\) )

∴ Y' = (2 , 3)

∵  (15 × \(\frac{1}{3}\) , 0 ×  \(\frac{1}{3}\) )

∴ Z' = (5, 0)

∴ The coordinates of the vertices of the image are (0, 0), (2, 3), (5, 0)

Answer:

ONL and MNL

Step-by-step explanation:

Adjacent angles are next to each other and share a side

ONL and MNL are next to each other and share a side

Answer:

x = 12.17

Step-by-step explanation:

we use the tan function

\(\tan \theta = \fra{opp}{ady}\\\\\tan(49) = \frac{14}{x}\\\\x=\frac{14}{\tan 49}\\\\x \approx 12.17\)

Answer:

\(d + q = 20\)

\(0.25d + 0.10q = 3.05\)

Step-by-step explanation:

Given

Coins = 20

Value = $3.05

Required

Determine the equation that represent this

From the question, we have that

d = the number of dimes

q = the number of quarters

This implies that;

\(d + q = 20\)

Also;

\(1 d=\$0.25\ \ and\ \\1 q= \$0.10\)---------- Standard unit of conversion;

This implies that

\(0.25d + 0.10q = 3.05\)

Hence, the equations are:

\(d + q = 20\)

\(0.25d + 0.10q = 3.05\)

The standard deviation for a binomial probability distribution is : √npq.

The correct option is (D) i.e., square root of npq

The standard deviation of a random variable, sample, statistical population, data set or probability distribution is the square root of its variance.

For a binomial distribution,

µ = np, which signifies the expected number of successes.

\(\sigma^2\) = npq , \(\sigma^2\) is the variance.

Since, the standard deviation is the square root of the variance,

Therefore, σ = Standard deviation = √npq

Thus, the standard deviation for a binomial probability distribution is given by √npq.

The correct option is (D)

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Answer:

196°

Step-by-step explanation:

A full circle has an arc measure of 360°.

360° - 164° = 196°

Answer: 196°

Statistical process control (SPC) is a technique used in quality control to monitor and control a process over time

What is used to periodically check that a process is in statistical control?

a. sampling

b. scrap parts

c. the process is only measured in the beginning 100 percent inspection.

a. Sampling is used to periodically check that a process is in statistical control.

Statistical process control (SPC) is a technique used in quality control to monitor and control a process over time. SPC involves collecting and analyzing data on the process, and using statistical methods to determine whether the process is in statistical control (i.e., producing consistent and predictable results) or is out of control (i.e., producing inconsistent or unpredictable results).

One way to monitor a process using SPC is to use sampling. This involves taking a sample of parts or products from the process at regular intervals, and measuring certain characteristics of the sample (such as dimensions, weight, or color). The data collected from the samples can then be analyzed using statistical methods to determine whether the process is in control or out of control.

If the data collected from the samples indicates that the process is out of control (i.e., producing inconsistent or unpredictable results), corrective action can be taken to bring the process back into control. By regularly monitoring and adjusting the process using SPC techniques like sampling, organizations can ensure that their processes are producing consistent and high-quality results.

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The polynomial : y = 3.6088 x³  - 0.3494 x² + 7.7644 x + 2.1870

Given,

Table

MATLAB  code :

clc; clear all; x = [1.15; -1.95; 2.15; -0.8; 0.1]; y = [16.42; -41.1; 53.06; -5.8; 2.52]; % taking cube of x p =x³; % taking square of x q = x²; %combining all columns X = [p q x];  % model of the form ax³ + bX² + cx + d % fit model in linear model mdl = fitlm(X,y); % printing coeffecients of matrix mdl.Coefficients

The table is attached below .

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Answer:

a = -9

Step-by-step explanation:

2(a+3) = -12

2a + 6 = -12

2a = -18

a = -9

Let's Check

2(-9 + 3) = -12

2(-6) = -12

-12 = -12

So, a = -9 is the correct answer.

Out of the 20,000 athletes, 788 can be expected to test positive for drugs during the Olympics.

During the Olympics, all athletes must pass a mandatory drug test administered by the International Olympic Committee before they are permitted to compete. Assuming a 1% drug use rate among 20,000 athletes, we can expect about 200 athletes to actually use drugs (1% of 20,000). With a 97% accurate drug test, 3% of the test results will be inaccurate.
Out of the 200 athletes using drugs, 97% will test positive, which equals 194 athletes (0.97 * 200). However, there are also 19,800 athletes not using drugs (20,000 - 200). Out of these, 3% will falsely test positive, which equals 594 athletes (0.03 * 19,800).
Therefore, approximately 788 athletes (194 + 594) can be expected to test positive for drugs during the Olympics.

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approximately probability is 194 athletes can be expected to test positive for drugs out of a total of 20,000 athletes.

What is Probability?

Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability was introduced in mathematics to predict how likely events are to occur.

To determine the approximate number of athletes expected to test positive for drugs out of a total of 20,000 athletes, we can calculate it based on the given accuracy rate of the drug test and the rate of drug use among athletes.

The rate of drug use among athletes is given as 1 athlete per 100, which can also be expressed as a probability of 1/100 or 0.01. This means that the probability of an athlete using drugs is 0.01.

The accuracy rate of the drug test is stated as 97%, which can be expressed as a probability of 0.97. This means that the probability of a drug test correctly identifying an athlete who is using drugs is 0.97

Now, we can calculate the expected number of athletes who will test positive for drugs using these probabilities.

Expected number of athletes testing positive = Total number of athletes * Probability of drug use * Probability of accurate drug test result

Expected number of athletes testing positive = 20,000 * 0.01 * 0.97

Expected number of athletes testing positive = 200 * 0.97

Expected number of athletes testing positive ≈ 194

Therefore, approximately probability is 194 athletes can be expected to test positive for drugs out of a total of 20,000 athletes.

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The gravitational potential energy of the stone-Earth system can be calculated before the stone is released, after it reaches the bottom of the well, and the change in gravitational potential energy during the process.

Gravitational potential energy is given by the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.

(a) Before the stone is released, it is held 11 m above the top edge of the well. The mass of the stone is 0.25 kg, and the acceleration due to gravity is approximately 9.8 m/s². Using the formula, the gravitational potential energy is calculated as PE = (0.25 kg)(9.8 m/s²)(11 m).

(b) After the stone reaches the bottom of the well, its height is 7.3 m. Using the same formula, the gravitational potential energy at this point is given by PE = (0.25 kg)(9.8 m/s²)(7.3 m).

(c) The change in gravitational potential energy can be determined by subtracting the initial potential energy from the final potential energy. The change in gravitational potential energy is equal to the gravitational potential energy after reaching the bottom of the well minus the gravitational potential energy before the stone was released.

By calculating these values, we can determine the specific numerical values for (a), (b), and (c) based on the given data.

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The probability that a valve was ground twice can be calculated using Bayes' theorem. Let G1 be the event that a valve is reground once and G2 be the event that a valve is reground twice. Then, the probability of a valve being scrapped given that it was reground once is 19%, and the probability of a valve meeting the specification given that it was reground twice is 100%. Therefore, using Bayes' theorem, we can calculate the probability of a valve being ground twice given that it was scrapped as (0.14 x 0.19) / ((0.14 x 0.19) + (0.24 x 0.81)) = 0.029. Thus, the probability that a valve was ground twice given that it was scrapped is 2.9%.

Bayes' theorem is a mathematical formula used to calculate conditional probabilities. It states that the probability of an event A given an event B is equal to the probability of event B given event A, multiplied by the probability of event A, divided by the probability of event B. In this problem, we want to calculate the probability of a valve being ground twice given that it was scrapped.
To apply Bayes' theorem, we first need to identify the relevant probabilities. We are given that after the first grind, 62% of the valves meet the specification, 24% are reground once, and 14% are scrapped. We are also given that of those valves that are reground, 81% meet the specification, and 19% are scrapped.
Let G1 be the event that a valve is reground once and G2 be the event that a valve is reground twice. Then, the probability of a valve being scrapped given that it was reground once is 19%. The probability of a valve meeting the specification given that it was reground twice is 100%, since all valves that are reground twice are guaranteed to meet the specification.
Using Bayes' theorem, we can calculate the probability of a valve being ground twice given that it was scrapped as follows:
P(G2|scrapped) = P(scrapped|G2) x P(G2) / P(scrapped)
where P(scrapped|G2) is the probability of a valve being scrapped given that it was reground twice, P(G2) is the probability of a valve being reground twice, and P(scrapped) is the probability of a valve being scrapped.
We already know that P(scrapped|G1) = 0.19, P(scrapped|G2) = 0, P(G1) = 0.24, P(G2) = (1 - 0.62 - 0.24 - 0.14) x P(G1) = 0.027, and P(scrapped) = 0.14.
Plugging in the values, we get:
P(G2|scrapped) = (0 x 0.027) / ((0.14 x 0.19) + (0.24 x 0.81)) = 0.029
Thus, the probability that a valve was ground twice given that it was scrapped is 2.9%.

In summary, we can use Bayes' theorem to calculate the probability of a valve being ground twice given that it was scrapped. We first identify the relevant probabilities, such as the probability of a valve being scrapped given that it was reground once or twice. We then apply Bayes' theorem to obtain the desired probability. In this case, the probability that a valve was ground twice given that it was scrapped is 2.9%.

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Step-by-step explanation:

\( \underline{ \underline{ \text{Given}}} : \)

\( \underline{ \underline { \text{To \: Find}}} : \)

\( \underline{ \underline{ \text{Solution}}} : \)

The new matrix obtained from a given matrix by interchanging it's rows and columns is called the transposition of matrix. It is denoted by \( \sf{ {A}^{T}} \). Again , Interchange it's rows and columns in order to find ' A '.

\( \tt{A = \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix}}\)

Now , LEFT HAND SIDE ( L.H.S )

\( \tt{ {A}^{2} - 5A+ 22I}\)

Here, I refers to identity matrix. A diagonal matrix in which all the elements of leading diagonal are 1 ( unit ) is called unit or identity matrix.

⟼ \(\begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} - 5 \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} + 22 \times \begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}\)

⟼ \(\begin{bmatrix} 2 \times 2 + 4 \times ( - 4)& 2 \times 4 + 4 \times 3 \\ - 4 \times 2 + 3 \times ( - 4) & - 4 \times 4 + 3 \times 3 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20& 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\)

⟼ \(\begin{bmatrix} 4 + ( - 16) & 8 + 12 \\ - 8 + ( - 12) & - 16 + 9 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\)

⟼ \( \begin{bmatrix} - 12 & 20\\ - 20& - 7 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\)

⟼ \(\begin{bmatrix} - 22 & 0 \\ 0& - 22 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\)

⟼ \(\begin{bmatrix} - 22 + 22 & 0 + 0 \\ 0 + 0 & - 22 + 22 \\ \end{bmatrix}\)

⟼ \(\begin{bmatrix} 0 & 0\\ 0 & 0 \\ \end{bmatrix}\)

⟼ \( \sf{0}\)

RIGHT HAND SIDE ( R.H.S ) : 0

L.H.S = R.H.S [ Hence , proved ! ]

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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Answer:

Step-by-step explanation:

\underline{ \underline{ \text{Given}}} :

\begin{gathered} \tt{ {A}^{T} = \begin{bmatrix} 2 & - 4 \\ 4 & 3 \\ \end{bmatrix}}\end{gathered}

\underline{ \underline { \text{To \: Find}}} :

\underline{ \underline{ \text{Solution}}} :

The new matrix obtained from a given matrix by interchanging it's rows and columns is called the transposition of matrix. It is denoted by \sf{ {A}^{T}}

. Again , Interchange it's rows and columns in order to find ' A '.

\begin{gathered} \tt{A = \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix}}\end{gathered}

Now , LEFT HAND SIDE ( L.H.S )

\tt{ {A}^{2} - 5A+ 22I}

Here, I refers to identity matrix. A diagonal matrix in which all the elements of leading diagonal are 1 ( unit ) is called unit or identity matrix.

⟼ \begin{gathered}\begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} - 5 \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} + 22 \times \begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered}\begin{bmatrix} 2 \times 2 + 4 \times ( - 4)& 2 \times 4 + 4 \times 3 \\ - 4 \times 2 + 3 \times ( - 4) & - 4 \times 4 + 3 \times 3 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20& 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered}\begin{bmatrix} 4 + ( - 16) & 8 + 12 \\ - 8 + ( - 12) & - 16 + 9 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered} \begin{bmatrix} - 12 & 20\\ - 20& - 7 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered}\begin{bmatrix} - 22 & 0 \\ 0& - 22 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered}\begin{bmatrix} - 22 + 22 & 0 + 0 \\ 0 + 0 & - 22 + 22 \\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered}\begin{bmatrix} 0 & 0\\ 0 & 0 \\ \end{bmatrix}\end{gathered}

⟼ \sf{0}0

RIGHT HAND SIDE ( R.H.S ) : 0