3 years ago

The radius of a circle is 6 cm. What is its diameter?

Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

5 0

Answer:

Step-by-step explanation:

d=2r=2, 6 =12

diameter=12

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Given the equations f(x) = x + 4 and d (x) = 2x + 5, find: f (1) + d (2)

Lina20 [59]

1 = x + 4
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2 = 2x + 5
3 = 2x
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3 years ago

Farmer Jones, and his wife, Dr. Jones, decide to build a fence in their field, to keep the sheep safe. Since Dr. Jones is a math

alex41 [277]

Answer: $\frac{8}{3}\times \sqrt{\frac{2}{5}}$

Step-by-step explanation:

Given two upward facing parabolas with equations

$y=6x^2 & y=x^2+2$

The two intersect at

$6x^2=x^2+2$

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$x^2$ = $\frac{2}{5}$

x= $\pm \sqrt{\frac{2}{5}}$

area enclosed by them is given by

A= $\int_{-\sqrt{\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left [ \left ( x^2+2\right )-\left ( 6x^2\right ) \right ]dx$

A= $\int_{\sqrt{-\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left ( 2-5x^2\right )dx$

A= $4\left [ \sqrt{\frac{2}{5}} \right ]-\frac{5}{3}\left [ \left ( \frac{2}{5}\right )^\frac{3}{2}-\left ( -\frac{2}{5}\right )^\frac{3}{2} \right ]$

A= $\frac{8}{3}\times \sqrt{\frac{2}{5}}$

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3 years ago

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Phoenix [80]

Answer:

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Answer:

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A) The solutions to this set of equation is where the graphs cross. They cross at point (-3, -2).

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C) These 2 functions cross at (4, 1). That is the solution.

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3 years ago