Ask question

Ask question

All categories

2 years ago

13

Show all of your work.

1 answer:

kirill [66]2 years ago

8 0

864 shirts :) 36x 24

You might be interested in

The graph represents function 1, and the equation represents function 2:

inessss [21]

Answer:

Option D.

Step-by-step explanation:

The slope of a horizontal line is 0.

It is given that the function 1 is a horizontal line that passing through the y-axis at y = 4.

It means the rate of change of function 1 is 0.

The slope intercept form of a linear function is 1

where, m is slope and b is y-intercept.

The function 2 is 2

On comparing (1) and (2), we get

The rate of change of function 2 is 8.

The difference between rate of change is

The rate of change of function 2 is 8 more than the rate of change of function 1.

Therefore, the correct option is D.

3 0

2 years ago

Read 2 more answers

What do the sides in a triangle add up to? I know that the inner angles of a triangle add up to 180°, but what do the sides of a

julia-pushkina [17]

Answer:

The most basic fact about triangles is that all the angles add up to a total of 180 degrees. The angle between the sides can be anything from greater than 0 to less than 180 degrees. The angles can't be 0 or 180 degrees, because the triangles would become straight lines.

Step-by-step explanation:

basic only

5 0

2 years ago

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro

Ugo [173]

Answer:

$Side\ B = 6.0$

$\alpha = 56.3$

$\theta = 93.7$

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with $\alpha, \beta, \theta$

[See Attachment for Triangle]

$A = 10cm$

$C = 12cm$

$\beta = 30$

What the question is to calculate the third length (Side B) and the other 2 angles ( $\alpha\ and\ \theta$ )

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

$B^2 = A^2 + C^2 - 2ABCos\beta$

Substitute: $A = 10$ , $C =12$ ; $\beta = 30$

$B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30$

$B^2 = 100 + 144 - 240*0.86602540378$

$B^2 = 100 + 144 - 207.846096907$

$B^2 = 36.153903093$

Take Square root of both sides

$\sqrt{B^2} = \sqrt{36.153903093}$

$B = \sqrt{36.153903093}$

$B = 6.0128115797$

$B = 6.0$ <em>(Approximated)</em>

Calculating Angle $\alpha$

$A^2 = B^2 + C^2 - 2BCCos\alpha$

Substitute: $A = 10$ , $C =12$ ; $B = 6$

$10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 36 + 144 - 144 *Cos\alpha$

$100 = 180 - 144 *Cos\alpha$

Subtract 180 from both sides

$100 - 180 = 180 - 180 - 144 *Cos\alpha$

$-80 = - 144 *Cos\alpha$

Divide both sides by -144

$\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}$

$\frac{-80}{-144} = Cos\alpha$

$0.5555556 = Cos\alpha$

Take arccos of both sides

$Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)$

$Cos^{-1}(0.5555556) = \alpha$

$56.25098078 = \alpha$

$\alpha = 56.3$ <em>(Approximated)</em>

Calculating $\theta$

Sum of angles in a triangle = 180

Hence;

$\alpha + \beta + \theta = 180$

$30 + 56.3 + \theta = 180$

$86.3 + \theta = 180$

Make $\theta$ the subject of formula

$\theta = 180 - 86.3$

$\theta = 93.7$

5 0

2 years ago

Ellis is painting wooden fenceposts before putti by then in his yard. They are each 5 feet tall and have a diameter of 1 foot. T

lawyer [7]

Answer:

39.25

Step-by-step explanation:

diameter. is 1 therefore radius is 1/2

formula is πr²h

where

h is height=5

=3.14x½²x5

=3.925 for one paint

therefore for 10 paints, we have

=3.925x10

=39.25

3 0

2 years ago

In the past, the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older

earnstyle [38]

Answer:

Step-by-step explanation:

In the past, mean of age of employees

i.e. $\mu = 40$

Recently sample was taken

n = sample size = 60

Mean of sample = 45

Std dev of sample s = 16

$H_0: \bar x = 40\\H_a: \bar x >40$

(Right tailed test)

Since only population std deviation is known we can use t test only

Std error = $\frac{s}{\sqrt{n} } \\=\frac{16}{\sqrt{61} } \\=2.049$

Mean difference = 45-40 =5

Test statistic t= $\frac{5}{2.049} \\=2.441$

df = 60

p value =0.008739

Since p < 0.05 we reject null hypothesis

The mean age has increased.

5 0

2 years ago