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tino4ka555 [31]
3 years ago
10

A small tree was planted at a height of 9 feet. The tree has been planted for 16 months and is now 57 feet tall. Write an equati

on to find how much it grew each month.
Mathematics
1 answer:
bekas [8.4K]3 years ago
5 0

Answer:

First you have to know what x and y mean in the equation. They told you in the problem that x is the avg # of feet the tree grew each month. That means that the number multiplied by x is the number of months.Y will be the total height. That number is 57.6. When y = 57.6, the number that should be multiplied by x is 16.The tree starts at 8 ft tall, not 0 feet, so the y-intercept in this equation should be 8, because even when 0 months have passed, it will be 8 ft tall. So you know the equation will have the +8 in it added to the slope times x part.It should be d. 16x + 8 = 57.6.

Step-by-step explanation:

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If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that
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Answer:

\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

<u>Given</u><u> </u><u>:</u><u>-</u>

• \sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}

<u>To</u><u> </u><u>Prove</u><u> </u><u>:</u><u>-</u><u> </u>

•\sf\implies cos2A =\dfrac{\sqrt3}{2}

<u>Proof </u><u>:</u><u>-</u><u> </u>

We know that ,

\sf\implies cos2A = 1 - 2sin^2A

Therefore , here substituting the value of sinA , we have ,

\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2

Simplify the whole square ,

\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}

Add the numbers in numerator ,

\sf\implies cos2A =  1-2\times \dfrac{4-2\sqrt3}{8}

Multiply it by 2 ,

\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}

Take out 2 common from the numerator ,

\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}

Simplify ,

\sf\implies cos2A =  1 -\dfrac{ 2-\sqrt3}{2}

Subtract the numbers ,

\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}

Simplify,

\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }

Hence Proved !

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