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Pavlova-9 [17]
3 years ago
7

What is the value of x? Enter your answer in the box. °

Mathematics
1 answer:
likoan [24]3 years ago
3 0

Answer:

111°

Step-by-step explanation:

7 sides

sum of angles= (180×7)-360

1260-360=900

158+126+125+121+139+120+x=900

789+x=900

x=900-789

x=111°

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$27.99 shoes ; 7 1/2 tax find the total cost or sale price to the nearest cent
Vinvika [58]
27.99x.075= 2.09925 or 2.10 Rounded
27.99+2.10= $30.09

5 0
3 years ago
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There are 16 marbles in a bowl: 3 are red, 6 are black, and 7 are green. Jimmy draws
tatiyna

Answer:

9/29 or about 31%

Step-by-step explanation:

6+3=9    6+3+7+16=29 so 9/29 (have a great time finishing the rest of your work) :)

7 0
3 years ago
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write an equation in slope-intercept form for a line that goes through (9,-2) and is parallel to y=7x+2​
lozanna [386]

Answer:

y=7x-65

Step-by-step explanation:

3 0
2 years ago
Find the area of the shaded portion in the equilateral triangle with sides 6. Show all work for full credit. (Hint: Assume that
Vadim26 [7]
So... if you notice the picture below

each circle, has their central angle at the vertex of the triangle
that simply means, 3 circles with a radius of 3, overlapping the triangle

now, the area in the middle, the shaded one, will be, the whole area of the triangle MINUS those 3 circle sectors

hmmmm each sector has 60°, that means, all three of them will then be 60+60+60 or 180°, so the area of those three sectors, can be combined into a 180° sector, well, hell, 180° is really half a circle

so.... the area of those three sectors of 60° each, all three combined, is the same area of half a circle with a radius of 3

so    \bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\textit{length of one side}\\\\
-----------------------------\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad r=radius\\\\
\textit{area of half a circle}\\\\
A=\cfrac{\pi r^2}{2}\\\\
-----------------------------\\\\

\bf \textit{now, let us use the side of 6, and radius of 3}
\\\\\\

\begin{array}{clclll}
\cfrac{6^2\sqrt{3}}{4}&-&\cfrac{\pi 3^2}{2}\\
\uparrow &&\uparrow \\
triangle's&&semi-circle's
\end{array}\impliedby \textit{area of shaded area}\\\\
-----------------------------\\\\
\boxed{\cfrac{36\sqrt{3}}{4}-\cfrac{9\pi }{2}}

you can, add the fractions if you want, or leave them like that, or get their difference by using their decimal format

8 0
3 years ago
Read 2 more answers
The line containing the median of the trapezoid whose vertices are R(-1, 5) , S(1, 8), T(7, -2), and U(2, 0).
bearhunter [10]

Answer:

y=-\dfrac{5}{3}x+6.5

Step-by-step explanation:

Given: trapezoid RSTU with vertices R(-1, 5), S(1, 8), T(7, -2), and U(2, 0).

Plot these points on the coordinate plane. As you can see, lines RU and ST are parallel lines and segments ST and RU are bases of the trapezoid.

The median of the trapezoid (the middle line) passes through the midpoints of the sides RS and TU. Find these midpoints:

M\left(\dfrac{x_S+x_R}{2},\dfrac{y_S+y_R}{2}\right)\Rightarrow M(0,6.5)\\ \\N\left(\dfrac{x_T+x_U}{2},\dfrac{y_T+y_U}{2}\right)\Rightarrow N(4.5,-1)

The equation of the line MN is

\dfrac{x-0}{4.5-0}=\dfrac{y-6.5}{-1-6.5}\\ \\\dfrac{x}{4.5}=\dfrac{y-6.5}{-7.5}\\ \\y=-\dfrac{5}{3}x+6.5

4 0
4 years ago
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