PLEASE HELP| calculate measure of angle H. round answer right nearest tenth of a degree

Answer:

8

Step-by-step explanation:

Explanation: The GCF of 56 and 64 is 8 , as 8 goes into 56 exactly 7 times and into 64 exactly 8 times. 8(8)+7(8)=8(7+8).

Answer: \(x=n\pi +(-1)^n\dfrac{\pi}{6}\), \(x=n\pi +(-1)^n\left(\dfrac{-\pi}{6}\right)\)

Step-by-step explanation:

Given

\(\sin x-\sqrt{1-3\sin^2x}=0\)

Take the square root term to the right-hand side

\(\Rightarrow \sin x=\sqrt{1-3\sin^2x}\\\text{Squaring both sides}\\\Rightarrow \sin^2x=1-3\sin^2x\\\Rightarrow 4\sin^2x=1\\\\\Rightarrow \sin^2x=\dfrac{1}{4}\\\\\Rightarrow \sin x=\pm\frac{1}{2}\)

for

\(\sin x=\dfrac{1}{2}\)

\(x=n\pi +(-1)^n\dfrac{\pi}{6}\)

for

\(\sin x=\dfrac{-1}{2}\)

\(x=n\pi +(-1)^n\left(\dfrac{-\pi}{6}\right)\)

Answer: the second one

Step-by-step explanation:

Answer:

In slope format, that would be y = 12x + 10

Step-by-step explanation:

The proof demonstrates that given a well-ordered set W, an isomorphic ordinal can be found using the function F. The uniqueness of this ordinal is established using the Replacement Axioms. The set F(W) is shown to exist for each x in W, and if the least F(W) exists, it serves as an isomorphism of VV onto -y.

Lemma 2.7: This is a previously stated lemma that is referenced in the proof. Unfortunately, without the specific details of Lemma 2.7, it's difficult to provide further explanation for its role in the proof.

Well-ordered set W: A well-ordered set is a set where every non-empty subset has a least element. In this proof, W is assumed to be a well-ordered set.

Isomorphic ordinal: An ordinal is a mathematical concept that extends the notion of natural numbers to represent order and magnitude. An isomorphic ordinal refers to an ordinal that has a one-to-one correspondence or mapping with another ordinal, preserving their order and magnitude properties.

Function F: The function F is defined to assign an ordinal o to each element x in W. This means that for every x in W, there is a corresponding ordinal o.

Existence and uniqueness: The proof asserts that if there exists an ordinal o that is isomorphic to a specific initial segment of the ordinal VV (the set of all ordinals), then this ordinal o is unique. In other words, there is only one ordinal that can be mapped to the initial segment of VV given by x.

Replacement Axioms: The Replacement Axioms are principles in set theory that allow the construction of new sets based on existing ones. In this case, the Replacement Axioms are used to assert that the set F(W) exists, which is the collection of all ordinals that can be assigned to elements of W.

For each x in W: The proof states that for every x in W, there exists an ordinal o that can be assigned to it. If there is no such ordinal, the proof suggests considering the least x for which such an ordinal does not exist.

The least F(W): The proof introduces the concept of the least element in the set F(W), denoted as the least F(W). If this least element exists, it serves as an isomorphism (a one-to-one mapping) of the set of all ordinals VV onto the ordinal -y.

Overall, the proof outlines the existence and uniqueness of an isomorphic ordinal that can be obtained from a well-ordered set W using the function F, and it relies on the Replacement Axioms and the concept of least element to establish this result.

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The correct expansion of Taylor series generated by the function f(x) will be \($$\begin{array}{r}=\frac{2 \pi}{5}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2-\frac{7}{12}(x-1)^3-\frac{7}{40}(x-5)^5 \\\text { first five non term }\}\end{array}$$\)

From the definition of Taylor series we know that the Taylor series of the function f at a is \($$\sum_{n=0}^\alpha \frac{f^n(a)}{n !}(n-a)^n=f(a)+f^{\prime}(a)(n-a)+\frac{f^{'''}(a)}{2 !}(n-a)^2$$\)

where \($f(x)=7 \tan ^{-1} x+\frac{\pi}{20}$\)

\($$\begin{aligned}f^{\prime}(x) & =\frac{7}{1+x^2} \\f^{\prime \prime}(x) & =\frac{7 \cdot(-1) 2 x}{\left(1+x^2\right)^2}=-\frac{14 x}{\left(1+x^2\right)^2} \\f^{\prime \prime \prime}(x) & =-\frac{\left(1+x^2\right)^2 \cdot 14-14 x \cdot 2\left(1+x^2\right) \cdot 2 x}{\left(1+x^2\right)^4} \\& =\frac{56 x^2\left(1+x^2\right)-14\left(1+x^2\right)^2}{\left(1+x^2\right)^4}\end{aligned}$$\)

\($\begin{aligned} & =\frac{56 x^2-14-14 x^2}{\left(1+x^2\right)^3}=\frac{42^2-14}{\left(1+x^2\right)^3} \\ f^N(x) & =\frac{\left(1+x^2\right)^3 \cdot 84 x-\left(42 x^2-14\right) 3\left(1+x^2\right)^2 \cdot 2 x}{\left(1+x^2\right)^6} \\ & =\frac{84 x\left(1+x^2\right)-6 x\left(42 x^2-14\right)}{\left(1+x^2\right)^4} \\ & =\frac{84 x+84 x^3-252 x^3+84 x}{\left(1+x^2\right)^4}=\frac{168 x-168 x^3}{\left(1+x^2\right)^4}\end{aligned}$\)

here a=1

\($$\begin{aligned}f(1) & =7 \tan ^{-1} 1+\frac{\pi}{20} \\& =\frac{7 \pi}{4}+\frac{\pi}{20}=\frac{35 \pi+\pi}{20}=\frac{36 \pi}{20}=\frac{9 \pi}{5} \\f^{\prime}(1) & =\frac{7}{1+1}=\frac{7}{2} \\f^{\prime \prime}(1) & =-\frac{14}{(1+1)^2}=-\frac{14}{4}=-\frac{7}{2} \\f^{\prime \prime \prime}(1) & =\frac{42-14}{(1+1)^3}=\frac{28}{8}=\frac{7}{2} \\f^{i v}(1) & =0\end{aligned}$$\)

\($\begin{aligned} f^{\prime v}(1) & =0 \\ f^v(x) & =\frac{\left(1+x^2\right)^4\left(168-504 x^2\right)-\left(168-168 x^3\right)}{4\left(1+x^2\right)^3 \cdot 2 x} \\ & =\frac{\left.\left(1+x^2\right)^8\right)\left(168-504 x^2\right)-8 x\left(168-168 x^3\right)}{\left(1+x^2\right)^5} \\ f^v(1) & =\frac{2(168-504)-8(168-168)}{2^5}\end{aligned}$\)

\(=-\frac{336}{24}=-\frac{42}{2}=-21\)

Taylor series polyromin of \($7 \tan ^{-1} x+\frac{\pi}{20}$\) is \($f(1)+f^{\prime}(1)(x-1)+\frac{f^{\prime \prime}(1)}{2 !}(x-1)^2+\frac{f^{\prime \prime \prime}(1)}{3 !}(x-1)^3$$$+\frac{f^{12}(1)}{4 !}(x-1)^4+\frac{f^2(1)}{5 !}(x-1)^5+\cdots \cdot$$\)

\($=\frac{9 \pi}{5}+\frac{7}{2}(x-1)+\frac{-\frac{7}{2}}{2 !}(x-1)^2+\frac{-\frac{7}{2}}{3 !}(x-1)^3$$+0+\frac{-21}{5 !}(x-5)^5$\)

first five non-zero term

\($=\frac{9 \pi}{5}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2-\frac{7}{12}(x-1)^3+0-\frac{21}{120}(x-5)^5$$=\frac{9 \pi}{5}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2-\frac{7}{12}(x-1)^3-\frac{7}{40}(x-5)^5$\)

(first five non zero term)

The correct question should be

Use the definition to find the first five nonzero terms of the Taylor series generated by the function \(f(x)=7 \tan ^{-1} x+\frac{\pi}{20}\) about the point  a=1

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Answer:

first option

Step-by-step explanation:

The perimeter is the distance around the composite shape, which is composed of a rectangle and a semi- circle.

P = 12 + 16 + 12 + πr ← ( rectangle and semi- circle )

   = 40 + (3.14 × 8)

    = 40 + 25.12

    = 65.12 in

----------------------------------

A = (16 × 12) + (0.5πr² ) ← rectangle and semi- circle

   = 192 + 0.5 × 3.14 × 64)

    = 192 + 100.48

    = 292.48 in²

Answer:

\( \frac{ \sqrt{2} }{2} \)

Step-by-step explanation:

I took the quiz :)

Answer:

  2.  $19,547.04

  3.  $10,276.54

  4.  $16,758.38

Step-by-step explanation:

You have three problems in compound interest with different initial payments, interest rates, and compounding intervals. You want the relationship between the initial payment and the future value.

The multiplier of a single payment earning annual interest rate r compounded n times per year for t years is ...

  k = (1 +r/n)^(nt)

For this problem, r=0.09, n=1, t=9. The multiplier of the payment is ...

  k = (1 +0.09/1)^(1·9) = 1.09^9 = 2.17189328

Then the future value of $9000 will be ...

  FV = $9000 × 2.17189328 ≈ $19,547.04

For this problem, r=0.10, n=12, t=4. The multiplier of the payment is ...

  k = (1 +0.10/12)^(12·4) = 1.08333...^48 = 1.48935410

Then the future value of $6900 will be ...

  FV = $6900 × 1.48935410 ≈ $10,276.54

For this problem, r=0.13, n=12, t=13. The multiplier of the payment is ...

  k = (1 +0.13/12)^(12·13) = 1.0108333...^156 = 5.3704484

Then the payment that gives a future value of $90,000 will be ...

  P = $90,000/5.3704484 = $16,758.38

__

Additional comment

For the spreadsheet calculation, we used the "Goal Seek" capability to adjust the value of cell F4 to 90000 by changing the value in cell B4.

We could have calculated the multiplier as above, then used it different ways for the different problems. Instead, we used one FV( ) function for all of the problems.

The domain of the given function is; (0, 2)

The range of the given function is; (-2, 6)

We want to evaluate the function f(x) = 4x - 2 for x = 0, 1 and 2.

Step 1;

f(0) = 4(0) - 2

f(1) = 4(1) - 2

f(2) = 4(2) - 2

Step 2;

f(0) = 0 - 2 = -2

f(1) = 4 - 2 = 2

f(2) = 8 - 2 = 6

The domain is the set of all possible input values which is (0, 2)

The range is the set of all possible output values which is (-2, 6)

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The quadratic function that models the vertical motion is h(t) = -16t² + 96t + 77

The maximum height is 221 ft

From the question, we are to write a quadratic function to model the vertical motion

h(t) = -16t² + v₀t + h₀

From the given information,

Initial velocity is 96 ft/s

That is, v₀ = 96 ft/s

and

Initial height is 77 ft

That is, h₀ = 77 ft

Thus,

The function becomes

h(t) = -16t² + 96t + 77

To determine the maximum height, we will compare the equation to the general form of a quadratic function

y = ax² + bx + c

t = -b/2a

Where t is the time taken to reach maximum height

By comparing with h(t) = -16t² + 96t + 77

a = -16 and b 96

Thus,

t = -96 / (2 × -16)

t = -96 / -32

t = 3 seconds

Now, we will determine the height at 3 seconds

h(t) = -16t² + 96t + 77

h(3) = -16(3)² + 96(3) + 77

h(3) = -16 × 9 + 288 + 77

h(3) = -144 + 288 + 77

h(3) = 221 ft

Hence, the maximum height is 221 ft

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The sprinter will complete the race in approximately 17.07 seconds.

To calculate the time for the race, we need to consider two parts: the acceleration phase and the constant velocity phase.

Acceleration Phase:

The acceleration of the sprinter is 1.64 m/s², and the distance covered during this phase is 25 m. We can use the equation of motion to calculate the time taken during acceleration:

v = u + at

Here:

v = final velocity (which is the velocity at the end of the acceleration phase)

u = initial velocity (which is 0 since the sprinter starts from rest)

a = acceleration

t = time

Rearranging the equation, we have:

t = (v - u) / a

Since the sprinter starts from rest, the initial velocity (u) is 0. Therefore:

t = v / a

Plugging in the values, we get:

t = 25 m / 1.64 m/s²

Constant Velocity Phase:

Once the sprinter reaches the end of the acceleration phase, the velocity remains constant. The remaining distance to be covered is 100 m - 25 m = 75 m. We can calculate the time taken during this phase using the formula:

t = d / v

Here:

d = distance

v = velocity

Plugging in the values, we get:

t = 75 m / (v)

Since the velocity remains constant, we can use the final velocity from the acceleration phase.

Now, let's calculate the time for each phase and sum them up to get the total race time:

Acceleration Phase:

t1 = 25 m / 1.64 m/s²

Constant Velocity Phase:

t2 = 75 m / v

Total race time:

Total time = t1 + t2

Let's calculate the values:

t1 = 25 m / 1.64 m/s² = 15.24 s (rounded to two decimal places)

Now, we need to calculate the final velocity (v) at the end of the acceleration phase. We can use the formula:

v = u + at

Here:

u = initial velocity (0 m/s)

a = acceleration (1.64 m/s²)

t = time (25 m)

Plugging in the values, we get:

v = 0 m/s + (1.64 m/s²)(25 m) = 41 m/s

Now, let's calculate the time for the constant velocity phase:

t2 = 75 m / 41 m/s ≈ 1.83 s (rounded to two decimal places)

Finally, let's calculate the total race time:

Total time = t1 + t2 = 15.24 s + 1.83 s ≈ 17.07 s (rounded to two decimal places)

Therefore, the sprinter will complete the race in approximately 17.07 seconds.

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Answer:

The cost of the game is $7 .

Step-by-step explanation:

By looking at the graph, the line at x = 4 is equals to y = 7

The measure of the central angle θ in degrees and then multiply by 1/(2π). Then that result by πr²

We will use the following points to find the area of the sector;

First of all, the angle that is formed in a full rotation is 2π.

The area of a circle is given by the formula; A = πr²

The central angle is given by the angle symbol θ. Thus, we have to multiply the area of the circle by a factor θ/2π

Thus, the area of a sector is given by the formula;

A = (θ/2π) * πr²

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Answer:

              vertex

ray

Step-by-step explanation:

The point where the rays intersect, upper right, is the vertex.

One side of the angle, bottom left, is a ray.

The triangle is a isosceles because there is two same sides and one different side

The value of Betty's earnings of $55,000 in 1992 dolars is $44,141.25.

The consumer price index measures the changes in price of a basket of good.

CPI = (cost of basket of goods in current period / cost of basket of goods in base period) x 100

Value of the current earnings in 1992 dollars = $55,000 / 1.246 = $44,141.25

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1. Mean = 33.61

2. Standard Deviation = 11.3284

3. Variance = 128.33

The given data are: \(22.8, 49.8, 17.5, 44.1, 33.2, 20.6, 43.2, 37.7\)

To get mean, we will sum up all the numbers and divide by the total count. The mean is:

= (22.8 + 49.8 + 17.5 + 44.1 + 33.2 + 20.6 + 43.2 + 37.7) / 8

= 268.9 / 8

= 33.62

Standard Deviation: \(\sqrt((22.8 - 33.61)^2 + (49.8 - 33.61)^2 + (17.5 - 33.61)^2 + (44.1 - 33.61)^2 + (33.2 - 33.61)^2 + (20.6 - 33.61)^2 + (43.2 - 33.61)^2 + (37.7 - 33.61)^2) / 8\)

= \(\sqrt{128.3336}\)

= 11.3284420818

= 11.3284

Variance = (Standard Deviation)^2

Variance = \(11.3284 ^2\)

Variance = 128.33264656

Variance = 128.33.

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Answer:

The amount of interest for this loan is 16.6%.

Step-by-step explanation:

Since Levi will make a purchase for $ 3000 with your credit card, you should explore the two options you have:

-On the one hand, a down payment of $ 1000, with fees of $ 125.

-on the other, a down payment of $ 1,300, with fees of $ 110.

In both cases, the fees are the same, and the interest too. Therefore, to determine the amount of installments and interest, the difference between the down payment of one option and the other must be taken and divided by the difference between the value of the installments of either option.

Thus, the difference of the down payment of 300 (1300 - 1000) must be divided by 15 (125 - 110). This yields a result of 20 (300/15), with which that will be the total of quotas until both have paid the same amount of money.

Thus, given that 1300 + 110 x 20 = 3,500, and that 1000 + 125 x 20 = 3,500, in both cases the total value to be paid is $ 3,500. Since 500 is 16.6% of 3000, the total purchase interest will be 16.6%.

Answer:

Number line D

Step-by-step explanation:

That line goes from 0 to -3 and subtracts 2, as the equation shows.

Answer:

y-intercept will be -4

Step-by-step explanation:

y=mx+b where 'b' indicates the y-intercept

Answer:

-6

Step-by-step explanation:

A. The following are sets A and B:

A = {2, 3, 5, 7, 11}

B = {1, 2, 3, 4, 6, 12}

C. Elements not in A or A' = ∅

B. The Venn diagram is attached

A universal set is a set which consists of all elements related to the given sets. It is denoted by U.

A. Set A:

A = {2, 3, 5, 7, 11}

Set B:

B = {1, 2, 3, 4, 6, 12}

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

A' = {1, 4, 6, 8, 9, 10, 12}

C. Elements not in A or A' = ∅

Complement of set A is refers to a set that contains the elements present in the universal set but not in set A.

Hence, the Venn diagram of sets A, B and U has been attached.

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Answer:

there is no real question, once you add one I'll be glad to help, but with this knowledge, I can't

Step-by-step explanation:

Step-by-step explanation:

Hey there!

Given equation is; 6x + y= 6.

6x + y -6=0........……(i)

Slope (m1) = (-coefficient of X)/(coefficient. of y)

Slope (m1) = (-6)/(1)

Therefore, slope of the equation is -6.

As per the condition of parallel lines, slope of 1st equation is equal to 2nd equation.

Therefore, m1=m2= -6

The another straight line passes through the point (2,-5). So to find the equation, use one-point formula.

(i.e (y-y1) = m2(x-x1)).

So, the equation is;

(y+5) = m2(x-2)

Now, we know that m1=m2= -6. Let's keep its value on it.

y+5 = -6(x-2)

Or, y+5 = -6x +12

Or, 6x+ y -7 = 0

Therefore, the equation which is parallel to 6x + y= 6 equation and passes through the point (2,-5) is 6x+y-7=0.

Hope it helps...

Answer:

550 coins

Step-by-step explanation:

2% = 2/100 = 0.02

11/0.02 = 550

The collection consists of 550 coins

Hope this helps

Answer:

25.2

Step-by-step explanation:

168 x .15 = 25.2

\(\large\text{Hey there!}\)

\(\rm{15\%\ of\ 168}\\\\\rm{= 15\% \times 168} \\\\\rm{= \dfrac{15}{100}\times 168}\\\\\rm{= \dfrac{15}{100}\times \dfrac{168}{1}}\\\\\rm{=\dfrac{15\times168}{100\times1}}\\\\\rm{= \dfrac{2,520}{100}}\\\\\rm{= 25.2}\)

\(\huge\text{Therefore, your answer should be: }\)

\(\huge\boxed{\frak{25.2}}\huge\checkmark\)

\(\huge\text{Good luck on your assignment \& enjoy your day!}\)

~\(\frak{Amphitrite1040:)}\)

Answer:

Below

Step-by-step explanation:

The equation to use for CONTINUOUS compounding is :

FV = PV e^(it)

18532.08 = 8327 * e^( i*10)        where i is the interest in decimal form

18532.08/8327  = e^(10i)

2.225541 = e^(10i)             now take natural log  (ln) of both sides to get

.8000 = 10i

i = .08     or 8% interest

The change represents growth and rate of increase is 4.2%

Exponential functions model a relationship in which a constant change in the independent variable gives the same proportional change.

Given that, an exponential function y = 970(1.042)ˣ

When the growth rate is greater than 1, then the function represents a growth.

The general form equation is:

y(x)= a(1+r)ˣ such that r is the growth percent.

So,

1+r = 1.042

r = 0.042

r = 4.2%

Hence, the growth rate is 4.2%

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