In transaction (A) on 5/8, the cash account is an

Bryan created a scale drawing of his rectangular bedroom. His drawing had dimensions of 6 inches by 8 inches. If the scale that

vlabodo [156]

Answer:

56 feet

Step-by-step explanation:

If 1 inch = 2 feet, we can first find the actual dimensions of his room by multiplying the number of inches by 2.

6(2)

= 12

8(2)

= 16

So, the actual dimensions of his room are 12 feet by 16 feet.

To find the perimeter, use the perimeter formula, P = 2l + 2w.

Plug in the length and width:

P = 2l + 2w

P = 2(12) + 2(16)

P = 24 + 32

P = 56

So, the perimeter of his actual room is 56 feet.

6 0

3 years ago

A regular hexagon has an apothem of 14.7 inches and a perimeter of 101.8 inches. What is the area of the hexagon?

Neporo4naja [7]

<u>Answer</u>

748.23 in²

<u>Explanation</u>

Apothem a line from the centre of a regular polygon at right angles to any of its sides. This can be said to be the height of the triangle in a polygon.

The base of the hexagon = 101.8 ÷ 6

= 19.96666..

= 19.9667 inches

Area of the hexagon = 1/2 × height × base length × 6

= 1/2 × 14.7 × 19.9667 × 6

= 748.23 in²

7 0

3 years ago

Read 2 more answers

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:

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