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saveliy_v [14]
4 years ago
12

50 students tried out for the baseball team. 35 students made the team find the percent of decrease

Mathematics
1 answer:
AleksAgata [21]4 years ago
7 0

Answer:

70% made it 30% didn't make it

Step-by-step explanation:

Step 1:

35/50 = x/100      Equation

Step 2:

50x = 3,500        Multiply

Step 4:

x = 3,500 ÷ 50      Divide

Answer:

x = 70

Hope This Helps :)

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A) 2-x,when x=5
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Answer:

a) 2-5 = -3

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

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 !

8 0
3 years ago
Find the midpoint of the segment with the following end points (10,5) and (6,9)
Taya2010 [7]

Answer:

The mid-point between the endpoints (10,5) and (6,9) is:

  • \left(x,\:y\right)=\left(8,\:7\right)

Step-by-step explanation:

Let (x, y) be the mid-point

Given the points

  • (10,5)
  • (6,9)

Using the formula to find the mid-point between the endpoints (10,5) and (6,9)

\left(x,\:y\right)=\left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)

Here:

\left(x_1,\:y_1\right)=\left(10,\:5\right),\:\left(x_2,\:y_2\right)=\left(6,\:9\right)

Thus,

\left(x,\:y\right)=\left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)

\left(x,\:y\right)=\left(\frac{6+10}{2},\:\frac{9+5}{2}\right)

\left(x,\:y\right)=\left(8,\:7\right)

Therefore, the mid-point between the endpoints (10,5) and (6,9) is:

  • \left(x,\:y\right)=\left(8,\:7\right)
3 0
3 years ago
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