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2 years ago

Please tell me how to work this out.

Mathematics

1 answer:

strojnjashka [21]2 years ago

5 0

Step-by-step explanation:

x³ - 4x² + 3x

= x(x² - 4x + 3)

= x(x -3)(x - 1)

× = 0, 1, 3

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Whats the answer??????

Temka [501]

4/1 x 3/2 = 12/2
6/9 x 1/7 = 6/63
5/6 x 2/1 = 10/6
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8/1 x 5/8 = 40/8

7 0

2 years ago

Read 2 more answers

Please help, I'm so confused! I will give brainliest!

nika2105 [10]

Hello,

1) Vertical

2) Supplementary

3) Corresponding

4) Alternate exterior

Hope it helps!

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3 years ago

Will mark as brainliest!

lora16 [44]

Answer:

Step-by-step explanation:

6 0

2 years ago

Find(f−g)(x) f(x)=x2−6xandg(x)=−2x+3 <br> A x2−4x+3 <br> B x2+4x−3 <br> C x2−4x−3 <br> D x2+4x+3

3241004551 [841]

Answer:

option C is the answer......

8 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:

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

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.

3 0

2 years ago