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2 years ago

6

The area of a rectangular wall of a barn is 144 square feet. Its length is 10 feet longer than the width. Find the length and wi

dth of the wall of the barn.
The width is __feet.

1 answer:

denpristay [2]2 years ago

3 0

Answer:

8 feet

Step-by-step explanation:

Let l and w be length and width, respectively. Then:

lw=144 and

l=w+10

So:

w(w+10)=144

w²+10w-144=0

(w+18)(w-8)=0

w=8 as the width ................

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3 years ago

9x2/3 help due in 3 min

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Answer:

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3 years ago

Please help me find the area of each shaded segment and round the answer to the nearest tenth :)

Keith_Richards [23]

Hi there!

In order to solve, we need to use the formula for the area of a circle, and the angle given to find the area of the sector. Then, subtract the area of the triangle to find the area of the segment. We would use the equation : A = x/360 (pi * r^2) - 1/2(b * h). We need to plug in 90 for the angle measurement, 8 for the radius, and 8 for both the base and height.

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3 0

3 years ago

Mason walked on Monday, Wensday, and Friday. Use these clues to find how far he walked each day. These distances were six-eights

krok68 [10]

Answer:

Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

Step-by-step explanation:

Given Mason walked on Monday, Wednesday and Friday. These distances were six-eights mile, one-fourth mile, and one-sixth mile. He did not walk the farthest on Monday. He walked less on Friday than Monday. we have to find how far he walked each day.

distances are $\frac{6}{8}, \frac{1}{4}, \frac{1}{6}$ that are 0.75, 0.25 and 0.17 respectively.

Now, he didn't walk farthest on Monday and also walked less on Friday than Monday.

∴ Less distance travelled is $\frac{1}{6}$ which is on friday and then $\frac{1}{4}$ on monday.

Rest distance which is $\frac{6}{8}$ on wednesday.

Hence, Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

4 0

3 years ago

Read 2 more answers

What is the equation of this circle in standard form?

creativ13 [48]

Answer:

Option C.

Step-by-step explanation:

Note : In the given points one point is (8,-2) instead of (-6,-2).

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$

where, (h,k) is center of circle and r is radius.

It is given that center of the circle is (-1,-2). So,

$h=-1,k=-2$

$(x-(-1))^2+(y-(-2))^2=r^2$

$(x+1)^2+(y+2)^2=r^2$ ...(1)

It is given that the circle passing through the point (8,-2),(-1,5),(6,-2),(-1,-9).

Substitute x=6 and y=-2 in equation (1).

$(6+1)^2+(-2+2)^2=r^2$

$(7)^2+(0)^2=r^2$

$49=r^2$

Substitute $r^2=49$ in equation (1).

$(x+1)^2+(y+2)^2=49$

Therefore, the correct option is C.

8 0

3 years ago