A _____ is a statement that has been proved

When the start fraction is 1 and the end fraction is 27, the value is a=-11/3.

A fraction is a part of a whole or, more broadly, any number of equal parts. In ordinary English, a fraction describes the number of components of a specific size and consists of a numerator displayed above a line (or before a slash like 12) and a non-zero denominator displayed below (or after) that line.

In addition to common fractions, numerators and denominators are used in rare fractions such as compound fractions, complex fractions, and mixed numerals.

Positive common fractions have natural integers as their numerator and denominator.

Here we need to find the value of n for which 9=(1/27)ᵃ⁺³

Now, we have to rewrite with a base of 3 we get,

3²=3⁻³.3ᵃ⁺³

2=-3(a+3)

2=-3a-9

=2+9=-3a

11=-3a

a=-11/3

Therefore, the values is a=-11/3

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Answer:

x= -2

Step-by-step explanation:

4x+4=-3x-10

7x+4=-10

7x=-14

x=-2    

Answer:

y= -4x-3

Step-by-step explanation:

First, pick two points from the graph.

For this problem I picked ( 0, -3 ) and ( -1, 1)

Next, find the slope

\(m=\frac{1+3}{-1-0}=\frac{4}{-1}=-4\)

Now since we can see from the graph that the y-intercept is -3, it is easy to place this function into slop-intercept form.

y= -4x-3

Answer:

Step-by-step explanation:

10, 14, 15, 16, 21, 27

Median: 15.5

Range: 17

Mode: None

Mean:17.7

If company has 2855 units in stock, then the probability of stockout is (b) 2.442%.

To calculate the probability of a stockout, we use the concept of the normal distribution. The historical demand average of 1447 units per week and a standard-deviation of 715 units, we assume that the demand follows a normal distribution.

To find the probability of a stockout, we determine how likely it is for the demand to exceed the current stock level of 2855 units.

First, we calculate the z-score, which measures the number of standard deviations the current stock level is away from the mean:

z = (2855 - 1447)/715 = 1.9818

Now, we find the probability of a stockout by calculating the area under the normal distribution curve to the right of this z-score.

This represents the probability of the demand exceeding the current stock level.

We know that probability corresponding to a z-score of 1.9818 is approximately 0.02442.

Therefore, the correct option is (b).

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Given

Mike took a taxi from home to Airport. The taxi driver charged $6 as initial fees and $3 per mile and the total fare is $24.

Required

we need to find total number of miles he covered on this ride.

Explanation

Let total number of miles he travelled be x

Then total fare =6+3x

i.e

So total number of miles covered is 6 miles.

The Taylor polynomial T3(x) of degree 3 for the function f(x) near x = 0 is given as: T3(x) = 4 - 3x + x^2 + 4x^3

A Taylor series is a series expansion of a function about a point. A one-dimensional Taylor series is an expansion of a real function f(x) about a point x=a is given by f(x)=f(a)+f^'(a)(x-a)+(f^('')(a))/(2!)(x-a)^2+(f^((3))(a))/(3!)(x-a)^3+...+(f^((n))(a))/(n!)(x-a)^n+.... .

If a=0, the expansion is known as a Maclaurin series.

Taylor's theorem (actually discovered first by Gregory) states that any function satisfying certain conditions can be expressed as a Taylor series.

The Taylor (or more general) series of a function f(x) about a point a up to order n may be found using Series[f,  {x, a, n}]. The nth term of a Taylor series of a function f can be computed in the Wolfram Language using SeriesCoefficient[f,  {x, a, n}] and is given by the inverse Z-transform  To find the values of f(0), f'(0), f''(0), and f'''(0), we need to differentiate T3(x) up to the third order and then evaluate the derivatives at x = 0.
So, let's start by finding the first derivative of T3(x):
T3'(x) = -3 + 2x + 12x^2
Now, we can evaluate T3(x), T3'(x), and T3''(x) at x = 0:
f(0) = T3(0) = 4 - 0 + 0 + 0 = 4
f'(0) = T3'(0) = -3 + 0 + 0 = -3
To find the second derivative, we differentiate T3'(x):
T3''(x) = 2 + 24x
Then, we evaluate T3''(x) at x = 0:
f''(0) = T3''(0) = 2 + 0 = 2
Finally, to find the third derivative, we differentiate T3''(x):
T3'''(x) = 24
And evaluate T3'''(x) at x = 0:
f'''(0) = T3'''(0) = 24
Therefore, the values of f(0), f'(0), f''(0), and f'''(0) for the function f(x) approximated near x = 0 by the 3rd degree Taylor polynomial T3(x) = 4 - 3x + x^2 + 4x^3 are:
f(0) = 4
f'(0) = -3
f''(0) = 2
f'''(0) = 24

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Answer:

-5/3

Step-by-step explanation:

Answer:

-2.4

-0.8

0.2

0.9

1.6

Step-by-step explanation:

I just know.

Answer:

3 1/2

Step-by-step explanation:

Hope this helped :)

Answer:

pretty sure it’s 35.6

Step-by-step explanation:

Answer:

11.35 g/mL

Step-by-step explanation:

D = m/v

D = 567.5 g ÷ 50.0 mL (I assume its measured in mL)

D = 11.35 g/mL

Answer:

11.35

Step-by-step explanation:

Answer:

5/42 or 0.1190

hope this helps :)

Answer:

6 1/2

Step-by-step explanation:

The difference between 13 9/20 and 6 7/8 can be found by converting both fractions to decimals and subtracting the two values.

13 9/20 can be expressed as a decimal as 13.45.

6 7/8 can be expressed as a decimal as 6.875.

So, the difference between 13 9/20 and 6 7/8 can be expressed as:

13.45 - 6.875 = 6.575

So the reasonable difference between 13 9/20 and 6 7/8 is 6.575.

To use differentials to estimate the value of 1.2√4, we need to first find a suitable function and a point close to the one we want to estimate. Since we are dealing with square roots, we can use the function f(x) = √x. The point closest to 4.2 (1.2 added to 4) that we know the exact value for is 4, as √4 = 2.

1. Find the derivative of f(x) = √x:
f'(x) = (1/2)x^(-1/2)

2. Evaluate the derivative at the known point, x = 4:
f'(4) = (1/2)(4)^(-1/2) = (1/2)(2)^(-1) = 1/4

3. Compute the differential, using Δx = 0.2 (the difference between 4.2 and 4):
Δy = f'(4)Δx = (1/4)(0.2) = 0.05

4. Estimate the value of 1.2√4 by adding Δy to the exact value at x = 4:
1.2√4 ≈ 2 + 0.05 = 2.05

Now, let's compare this to the exact value of 1.2√4:
Exact value: 1.2√4 = 1.2(√4.2) ≈ 1.2(2.0494) ≈ 2.0593

The estimated value using differentials is 2.05, while the exact value is approximately 2.0593. The two values are relatively close, demonstrating the effectiveness of using differentials for estimation.

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Answer:

Step-by-step explanation:

tanx + cotx = secx / sinx

take left hand side

=tanx + cotx

=sinx/cosx + cosx/sinx

=sinx*sinx+cosx*cosx/sinx*cosx

=sin^2x + cos^2x/sinx*cosx

=1/sinx*cosx

=1/cosx * 1/sinx

=secx/sinx

at second last step i changed multiplication sign into divide sign and write the reciprocal of 1/sinx which is sinx.

tried my best to show the steps!! all i did was isolate to find the missing widths attached to the respective lengths (divided surface area by width) and then multiply both sides by two and add them together to find the perimeter.

The dimensions of the garden that would require the least amount of fencing are 6 ft in length and 6 ft in width.

The dimensions are the sides of a geometric shape.

The lengths of the three gardens are given as 6 ft, 9 ft, and 12 ft respectively.

If the area of the garden is 36 feet squared, then the widths for the three lengths will be as follows:

The width of the garden with a length of 6 ft \(= \dfrac{36}{6} = 6\ ft\)

The width of the garden with a length of 9 ft \(= \dfrac{36}{9} = 4\ ft\)

The width of the garden with a length of 6 ft \(= \dfrac{36}{12} = 3\ ft\)

The parameters are calculated as:

The parameter of the garden with dimensions (6 ft, 6 ft) = 2(6 + 6) = 24 ft

The parameter of the garden with dimensions (9 ft, 4 ft) = 2(9 + 4) = 26 ft

The parameter of the garden with dimensions (12 ft, 3 ft) = 2(12 + 3) = 30 ft

The least of the parameter is of the garden with dimensions (6 ft, 6 ft).

Thus, the garden with dimensions (6 ft, 6 ft) would require the least amount of fencing.

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Step-by-step explanation:

x = number of years after 2007.

x = 0 for 2007.

PC(x) is a function that calculates how many pounds of chicken the average adult will eat every year (x) after 2007 (up to 2012).

PC(x) = 0.8x + 52

0 <= x <= 5

in 2007 (x = 0) the average adult ate 52 pounds.

in every year after 2007 until 2012 0.8 pounds get added to the amount of the previous year.

so, in 2012 (x = 5) the average adult will eat

PC(5) = 0.8×5 + 52 = 4 + 52 = 56 pounds of chicken.

F(4)4 is the x value 4. Find where the line is on th y axis at x = 4

The dot is between 60 and 70 so the answer would be 64

Answer: D. 64

Answer:

the answer is d

Step-by-step explanation:

if you look on the graph where the x-value equals 4, the y-value of that point is your answer and d is the only value in that range

Answer

1500/60=25

$25

We have to calculate the relation between the perimeter of similar triangles.

As the perimeter is the sum of the lengths of the sides, we can demonstrate that the relation between perimeters is equal to the scale factor between sides.

In this case, the scale factor is k=2/5.

The perimeter of the left triangle is:

and the perimeter of the triangle at the right is:

If we write the ratio, we get:

Answer: 2/5

, the correct answers are:

Orientation (North Arrow)

Scale

Legend

Neatline Border

The four primary cartographic elements are:

Orientation (North Arrow): This element indicates the orientation or direction of the map, typically pointing towards the north. It helps users understand the map's alignment and relation to the real world.

Scale: The scale provides a ratio or representative fraction that shows the relationship between distances on the map and corresponding distances on the Earth's surface. It helps users determine the actual size or distance of features on the map.

Legend: The legend, also known as a key, explains the symbols, colors, and other graphic elements used on the map. It helps users understand the meaning or representation of various features or data.

Neatline Border: The neatline border defines the outer boundary of the map. It is a solid line that encloses the map area and separates it from the surrounding space or background.

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A. \(\frac{v}{3} = 20 -v\\\)

B. \(3(n +4) = 28\)

Answer:

Step-by-step explanation:

Let,

The missing length be = x yd

As we know According to Pythagoras Theorem,

\( {50yd}^{2} = {30yd}^{2} + {x \: yd}^{2} \)

\( = > 2500 = 900 + {x}^{2} \)

\( = > {x}^{2} = 2500 - 900\)

\( = > {x}^{2} = 1600\)

\( = > x = \sqrt{1600} = \sqrt{40 \times 40} \)

=> x = 40

Hence,

Missing length will be 40 yd (Ans)

Answer:

C. 4 times x squared plus 6 times x minus 3 minus 12 over the quantity 2 times x plus 1

Step-by-step explanation:

f(x) = 8x^3 + 16x^2 − 15

g(x) = 2x + 1.

f(x) / g(x) = (8x^3 + 16x^2 − 15) / (2x + 1)

Using long division

Answer:

4 times x squared plus 6 times x minus 3 minus 12 over the quantity 2 times x plus 1

Step-by-step explanation:

i did the test

Answer:

Option (B)

Step-by-step explanation:

By adding two polynomials,

(5xy² + 3x² - 7) + (3x²y² - xy² + 3y² + 4)

Now by adding the similar terms,

= (5xy² - xy²) + 3x² + 3x²y² + 3y² - 7 + 4

= 4xy²+ 3x² + 3x²y² + 3y² - 3

= 3x²y² + 4xy² + 3x² + 3y² - 3

Therefore, Option (B) will be the correct option.

To test the difference between two independent population means with two samples, you need to calculate the degrees of freedom (df). Here's a step-by-step explanation:

1. Identify the sample sizes: The first sample has a size of 28 (n1 = 28), and the second sample has a size of 27 (n2 = 27).

2. Calculate the degrees of freedom for each sample: For each sample, subtract 1 from the sample size. For sample 1, df1 = n1 - 1 = 28 - 1 = 27. For sample 2, df2 = n2 - 1 = 27 - 1 = 26.

3. Combine the degrees of freedom: Add the degrees of freedom from each sample together to get the total degrees of freedom: df = df1 + df2 = 27 + 26 = 53.

In this case, you will use 53 degrees of freedom to find the critical value for testing the difference between the two independent population means.

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To write each vector in trigonometric form, we need to express them in terms of magnitude and angle.

18. \(\( \mathbf{b} = (\sqrt{19}, -4) \)\)

The magnitude of vector \(\( \mathbf{b} \) is \( \sqrt{(\sqrt{19})^2 + (-4)^2} = \sqrt{19 + 16} = \sqrt{35} \).\)

The angle of vector \(\( \mathbf{b} \)\) with respect to the positive x-axis can be found using the arctan function:

\(\( \mathbf{b} \) is \( \sqrt{35} \, \text{cis}(\arctan\left(\frac{-4}{\sqrt{19}}\right)) \).\)

So, the trigonometric form of vector \(\( \mathbf{b} \) is \( \sqrt{35} \, \text{cis}(\arctan\left(\frac{-4}{\sqrt{19}}\right)) \).\)

19. \(\( \mathbf{r} = 16i + 4j \)\)

The magnitude of vector \(\( \mathbf{r} \) is \( \sqrt{(16)^2 + (4)^2} = \sqrt{256 + 16} = \sqrt{272} = 16\sqrt{17} \).\)

The angle of vector \(\( \mathbf{r} \)\) with respect to the positive x-axis is 0 degrees since the vector lies along the x-axis.

So, the trigonometric form of vector \(\( \mathbf{r} \) is \( 16\sqrt{17} \, \text{cis}(0^\circ) \).\)

20.  \(\( \mathbf{k} = 4\sqrt{2}i - 2j \)\)

The magnitude of vector \(\( \mathbf{k} \) is \( \sqrt{(4\sqrt{2})^2 + (-2)^2} = \sqrt{32 + 4} = \sqrt{36} = 6 \).\)

The angle of vector \(\( \mathbf{k} \)\) with respect to the positive x-axis can be found using the arctan function:

\(\( \theta = \arctan\left(\frac{-2}{4\sqrt{2}}\right) \)\)

So, the trigonometric form of vector \(\( \mathbf{k} \) is \( 6 \, \text{cis}(\arctan\left(\frac{-2}{4\sqrt{2}}\right)) \).\)

21. \(\( \overrightarrow{CD} \) with C(2, 10) and D(-3, 8)\)

To find the vector \(\( \overrightarrow{CD} \)\), we subtract the coordinates of point C from the coordinates of point D:

\(\( \overrightarrow{CD} = \langle -3 - 2, 8 - 10 \rangle = \langle -5, -2 \rangle \)\)

The magnitude of vector \\(( \overrightarrow{CD} \) is \( \sqrt{(-5)^2 + (-2)^2} = \sqrt{29} \).\)

The angle of vector \(\( \overrightarrow{CD} \)\) with respect to the positive x-axis can be found using the arctan function:

\(\( \theta = \arctan\left(\frac{-2}{-5}\right) = \arctan\left(\frac{2}{5}\right) \)\)

So, the trigonometric form of vector \(\( \overrightarrow{CD} \) is \( \sqrt{29} \, \text{cis}(\arctan\left(\frac{2}{5}\right)) \).\)

22. overnighter \({TU} \) with T(-3, -4) and U(3, 8)\)

To find the vector we subtract the coordinates of point T from the coordinates of point U:

\(\( \overrightarrow{TU} = \langle 3 - (-3), 8 - (-4) \rangle = \langle 6, 12 \rangle \)\)

The magnitude of vector \(\( \overrightarrow{TU} \) is \( \sqrt{(6)^2 + (12)^2} = \sqrt{36 + 144} = \sqrt{180} = 6\sqrt{5} \).\)

The angle of vector \(\( \overrightarrow{TU} \)\) with respect to the positive x-axis can be found using the arctan function:

\(\( \theta = \arctan\left(\frac{12}{6}\right) = \arctan(2) \)\)\(\( \overrightarrow{TU} \),\)

So, the trigonometric form of vector \(\( \overrightarrow{TU} \) is \( 6\sqrt{5} \, \text{cis}(\arctan(2)) \).\)

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