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2 years ago

What is the area of the circle?

Mathematics

1 answer:

hodyreva [135]2 years ago

4 0

Diameter = radius x 2

So, the radius is half of the denominator.

10 / 2 = 5

A = pi x r^2

A = pi x 5^2

A = 3 x 25

A = 75 ft^2

Hope this helps!

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Item 6 Graph △JKL and its image after a reflection in the y-axis. J(5, 3), K(1,−2), L(−3, 4)

posledela

Answer:

J(-5,3) K(-1,-2) L(3,4)

Step-by-step explanation:

Reflections over the y axis make the x its opposite. 5 would change to -5, 1 would change to -1, and -3 would change to 3.

7 0

2 years ago

Solve for x in the equation x^2 + 2x+ 1 = 17.

muminat

Answer:

first, subtract 17 on both sides: x²+2x-16=0

this cannot be factored, so use the quadratic formula to solve for x:

b²-4ac=2²-4(10(-16)=4+64=68

√68=2√17

so x=(-2+2√17)/2 or x=(-2-2√17)/2

x=-1+√17 or x=-1-√17

Step-by-step explanation:

7 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of

marusya05 [52]

The Lagrangian is

$L(x,y,\lambda)=x+8y+\lambda(x^2+y^2-4)$

It has critical points where the first order derivatives vanish:

$L_x=1+2\lambda x=0\implies\lambda=-\dfrac1{2x}$

$L_y=8+2\lambda y=0\implies\lambda=-\dfrac4y$

$L_\lambda=x^2+y^2-4=0$

From the first two equations we get

$-\dfrac1{2x}=-\dfrac4y\implies y=8x$

Then

$x^2+y^2=65x^2=4\implies x=\pm\dfrac2{\sqrt{65}}\implies y=\pm\dfrac{16}{\sqrt{65}}$

At these critical points, we have

$f\left(\dfrac2{\sqrt{65}},\dfrac{16}{\sqrt{65}}\right)=2\sqrt{65}\approx16.125$ (maximum)

$f\left(-\dfrac2{\sqrt{65}},-\dfrac{16}{\sqrt{65}}\right)=-2\sqrt{65}\approx-16.125$ (minimum)

5 0

3 years ago

Simpify the expression. -2(y + 3) -7y

jenyasd209 [6]

-2y-6-7y
-9y-6
I hope that this helps!

4 0

2 years ago

Which statement about the equation is true

8090 [49]

They equal the same so I’m pretty sure it has no solution x

6 0

3 years ago

Read 2 more answers