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3 years ago

3.5 hours; $10 What is the unit rate per hour?

Mathematics

2 answers:

Nana76 [90]3 years ago

4 0

$2.86/h

To find the unit rate, divide $10 by 3.5 hours

andriy [413]3 years ago

3 0

The answer is $2.86 because it isthat much per hour

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Find the slope in (-2,8) and (1,6)

Salsk061 [2.6K]

Formula for slope: $\frac{y2 - y1}{x2 - x1}$

y₂: 6

y₁: 8

x₂: 1

x₁: -2

Plug them into the slope equation:

$\frac{6 - 8}{1 - (-2)}$

--> $\frac{-2}{1 + 2}$

---> $-\frac{2}{3}$

The slope is $-\frac{2}{3}$ .

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3 years ago

Which of the following is the correct graph of the compound inequality 4p + 1 > −7 or 6p + 3 < 33?. . A.a number line with

r-ruslan [8.4K]

We are given compound inequality 4p + 1 > −7 or 6p + 3 < 33.

Let us solve each of the inequality one by one.

4p + 1 > −7

Subtracting 1 from both sides, we get

4p + 1-1 > -7-1

4p > -8

Dividing both sides by 4, we get

p > -2. <em> (Shading right side for greater than sign)</em>

Solving 6p + 3 < 33.

Subtracting 3 from both sides, we get

6p + 3-3 < 33-3.

6p < 30

Dividing both sides by 6, we get

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<em>We have less than and greater than symbols in both inequalities, therefore we would have open circles(dots) on -2 and 5.</em>

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4 years ago

A 140 kg camera is suspended by two wires over a 40 metre wide football field to get shots of the action from above. At one poin

Katen [24]

Answer:

Please red the answer below

Step-by-step explanation:

In order to determine the length of each cable you use the Newton second law for each component of the forces involved in the situation.

For the x component you have:

$T_1cos\theta_1-T_2cos\theta_2=0$ (1)

T1: tension of the first cable = 1500N

T2: tension of the second cable = 800N

θ1: angle between the horizontal and the first cable

θ2: angle between the horizontal and the first cable

For the y component you have:

$T_1sin\theta_1+T_2sin\theta_2-W=0$ (2)

W: weight of the camera = Mg = (140kg)(9.8m/s^2) = 1372N

You can squared both equations (1) and (2) and the sum the two equations:

$T_1^2cos^2\theta_1=T_2^2cos^2\theta_2\\\\T_1^2sin^2\theta_1=T_2^2sin^2\theta_2-2WT_2sin\theta_2+W^2$

Then, you sum the equations:

$T_1^2(cos^2\theta_1+sin^2\theta_1)=T_2^2(sin^2\theta_2+cos^2\theta_2)-2Wsin\theta_2+W^2$ (3)

Next, you use the following identity:

$sin^2\theta+cos^2\theta=1$

and you obtain in the equation (3):

$T_1^2=T_2^2-2WT_2sin\theta_2+W^2\\\\sin\theta_2=\frac{T_2^2-T_1^2+W^2}{2WT_2}=\frac{(800N)^2-(1500)^2+(1372N)^2}{2(800N)(1372N)}=0.066\\\\\theta_2=sin^{-1}0.066=27.23\°$

With this values you can calculate the value of the another angle, by using the equation (1):

$\theta_1=cos^{-1}(\frac{T_2cos(27.23\°)}{T_1})=cos^{-1}(\frac{(800N)(cos27.23\°)}{1500N})\\\\\theta_1=61.69\°$

Now, you can calculate the length of each cable by using the information about the width of the football field. You use the following trigonometric relation:

$l_1cos\theta_1=40-d\\\\l_2cos\theta_2=d\\\\$

d: distance to the right side of the field

By using the cosine law you can fins a system of equation and then you can calculate the values of l1 and l2.

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3 years ago