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3 years ago

Plzz help i will give brainliest plzzzz

Mathematics

2 answers:

lisabon 2012 [21]3 years ago

8 0

B is the answer
At least that’s what my teacher said.

Rashid [163]3 years ago

6 0

Answer:

Step-by-step explanation:

B gives you 8/12, which can be simplified down to 2/3.

You might be interested in

Slove the inequality 8(1/2x-1/4)>12-2x

HACTEHA [7]

8(1/2x-1/4)>12-2x

4x-2>12-2x

6x>14

x>7/3

Hope this helps! ;)

6 0

4 years ago

How do you even do this?! i’m so confused

SIZIF [17.4K]

Answer:

Step-by-step explanation:

Pretend the y-axis is a wall, on the left side it is decreasing and on the right it is increasing, so A. and B. aren't correct. C. and D. are left, x<0 means on the left side of the y-axis because x (any number) is less than 0 and on the left side it is decreasing so C. isn't correct so D. is your answer because x>0 means x is greater than 0 so it's on the left side of the y-axis.

I hope this helps!

7 0

2 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:

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