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Dvinal [7]
3 years ago
5

Evaluate the function: f(x)=4x−2, find f(8)

Mathematics
1 answer:
valina [46]3 years ago
4 0

Answer:

f (8) = 30

Step-by-step explanation:

f(x)=4x-2\\\\f(8)=4(8)-2\\\\f(8)=32-2\\\\\boxed{f(8)=30}

Hope this helps.

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The volume of a prism is 48 cubic centimeters. If the length is 2 centimeters and the width is 4 centimeters. What is the height
Pie

Answer:

6

Step-by-step explanation:

volume is (lwh) since youre trying to find the height just multiply the 2 and 4 together to get 8. Then divide 8 by 48 to get your height.

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NEED HELP! The endpoints of WX are W(2,-7) and X(5,-4).
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Answer:

3√2 or 4.24

Step-by-step explanation:

d(w,x) = √(5-2)² + (-4-(-7))² = √18 = 3√2               (4.24)

3 0
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Forms of money in the United States consist of paper money, coins, and _____.
Maurinko [17]
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5 0
3 years ago
Read 2 more answers
PLS HELP
Umnica [9.8K]
The answer is 60, 60, 120, 120
do you go to rsm for geometry on saturday because i think i know you and this is exactly like the geometry homework 
8 0
3 years ago
Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.
sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

  1. All the sides of the rhombus are congruent:  |DG|\cong |GF| \cong |EF| \cong |DE|
  2. The diagonals are perpendicular

Using the distance formula; d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}

\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}

|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}

\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}

|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}

\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}

|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}

\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}

Using the slope formula; m=\frac{y_2-y_1}{x_2-x_1}

The slope of EG is m_{EG}=\frac{2c-0}{0-0}

\implies m_{EG}=\frac{2c}{0}

The slope of EG is undefined hence it is a vertical line.

The slope of  DF is m_{DF}=\frac{c-c}{a+b-(-a-b)}

\implies m_{DF}=\frac{0}{2a+2b)}=0

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since |DG|\cong |GF| \cong |EF| \cong |DE| and DF \perp FG , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is m_{DE}=\frac{2c-c}{0-(-a-b)}

m_{DE}=\frac{c}{a+b)}

The slope of FG is m_{FG}=\frac{c-0}{a+b-0}

\implies m_{FG}=\frac{c}{a+b}

5 0
3 years ago
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