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3 years ago

Geometry IXL help pls !

Mathematics

1 answer:

Trava [24]3 years ago

3 0

Answer:

The answer to your question is 10 in

Step-by-step explanation:

Data

Circumference = 20π in

Radius = x

Formula

Circumference = πd

Process

1.- Equal the formula to the circumference

πd = 20π

2.- Solve for d

d = 20π / π

3.- Simplification

d = 20

4.- Find the radius

r = d/2

r = 20/2

r = 10 in

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The shoes were $175. They are now 10% off. How much are they now?

kondor19780726 [428]

Sale Price = $157.5 (answer). This means the cost of the item to you is $157.5. You will pay $157.5 for a item with original price of $175 when discounted 10%. In this example, if you buy an item at $175 with 10% discount, you will pay 175 - 17.5 = 157.50 dollars.

Hope this helps!

3 0

3 years ago

Find the equation of the line that passes through the points : (4 , -1 ) and (0 , -3)

Sergio [31]

Answer:

B. y= 1/2 x - 3

Step-by-step explanation:

base equation: y= mx+ b

slope(m) equation: $\frac{y_{2}-y_{2} }{x_{2} -x_{1} }$

y2= -1 y1= -3

x2= 4 x1= 0

$\frac{-1 - (-3)}{4-0} =\frac{2}{4} = \frac{1}{2}$

y- intercept: -3

found by taking the y from (0,-3)

3 0

3 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

3 years ago

I think I have the answer but can anyone help?

kodGreya [7K]

Answer:

d 21.94

Step-by-step explanation:

11.99×4.99+2.09+2.09+.78=21.94

8 0

4 years ago

Based on its number of terms, what is the polynomial 2p2 – 2p called? PLS HELP

fenix001 [56]

Answer: quadratic

Step-by-step explanation:

3 0

3 years ago