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4 years ago

Which segment is NOT A RADIUS?????????????

Mathematics

1 answer:

Minchanka [31]4 years ago

4 0

Answer:

Step-by-step explanation:

A radius is from the center of the circle to the outside edge

LM does not go from the center of the circle

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If y=6 when x=2/3 find x when y=12

motikmotik

Y=6 x=2/3
y=12 x=?
6*2=12, 2/3*2=?
?=2*2/3
?=1 and 1/3

7 0

3 years ago

Read 2 more answers

Kirsten and Molly are on a 3-day bike ride. The entire route is 156 miles. On the first day they biked 64 miles, and on the seco

timama [110]

The distance that Kirsten and Molly need to bike on their third day is obtained by subtracting the sum of the distances they have traveled after two days from the length of the entire route.

distance left = 156 miles - (64 miles + 45 miles)
= 47 miles
Thus, Kirsten and Molly need to bike 47 miles on their third day.

3 0

4 years ago

6(1/2w-3/4 what does this equal to

Delicious77 [7]

6(1/2w-3/4)

distribute

6*1/2 w - 6 * 3/4

3 w -18/4

divide top and bottom by 2

3w -9/2

change to a mixed number (9/2 = 2 with 1 left over)

3w -4 1/2

5 0

3 years ago

Hey can anyone help me with this question please and thank you !

Inessa [10]

Answer:

(6x-12)+(8x-4)=180

14x-16=180

14x=196

x=14

Step-by-step explanation:

3 0

3 years ago

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. Y = (4/9) x

frez [133]

Answer:

V = 8.06 cubed units

Step-by-step explanation:

You have the following curves:

$y_1=\frac{4}{9}x^2=f(x)\\\\y_2=\frac{13}{9}-x^2=g(x)$

In order to calculate the solid of revolution bounded by the previous curves and the x axis, you use the following formula:

$V=\pi \int_a^b [(g(x))^2-(f(x))^2]dx$ (1)

To determine the limits of the integral you equal both curves f=g and solve for x:

$f(x)=g(x)\\\\\frac{4}{9}x^2=\frac{13}{9}-x^2\\\\\frac{4}{9}x^2+x^2=\frac{13}{9}\\\\\frac{13}{9}x^2=\frac{13}{9}\\\\x=\pm 1$

Then, the limits are a = -1 and b = 1

You replace f(x), g(x), a and b in the equation (1):

$V=\pi \int_{-1}^{1}[(\frac{13}{9}-x^2)^2-(\frac{4}{9}x^2)^2]dx\\\\V=\pi \int_{-1}^1[\frac{169}{81}-\frac{26}{9}x^2+x^4-\frac{16}{81}x^4]dx\\\\V=\pi \int_{-1}^1 [\frac{169}{81}-\frac{26}{9}x^2+\frac{65}{81}x^4]dx\\\\V=\pi [\frac{169}{81}x-\frac{26}{27}x^3+\frac{65}{405}x^5]_{-1}^1\\\\V\approx8.06\ cubed\ units$

The volume of the solid of revolution is approximately 8.06 cubed units

8 0

4 years ago