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snow_lady [41]
2 years ago
10

Hello can anyone help me with these problems please and thank you

Mathematics
1 answer:
Schach [20]2 years ago
5 0

Answer:

94)=154°

95)=40°

96)=11

97)=11

98)=9

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9.56? sorry if it is wrong <3

Step-by-step explanation:

6 0
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Can you please help me do this?
nlexa [21]

Answer: U=39

Step-by-step explanation:

6=u -15/4

cross multiply ✖

U-15=6*4

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2 years ago
Write an equation for a line perpendicular to y = -5x – 4 and passing through the point (15,5)
stiv31 [10]

Answer:

y=1/5x+2

Step-by-step explanation:

In perpendicular lines the slopes are negative reciprocals so the slope of other line is 1/5x

And they intersect at (15,5) to find the y intercept(b)

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Using a number line what is the equivalent fraction to 1/4
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Recall that in a 30 – 60 – 90 triangle, if the shortest leg measures x units, then the longer leg measures xStartRoot 3 EndRoot
Brums [2.3K]

The area of the shaded region is  \rm (150\sqrt{3} \ - 75\pi ) \ feet^2 option first is correct.

It is given that a circle is inscribed in a regular hexagon with sides of 10 feet.

It is required to find the shaded area (missing data is attached shown in the picture).

<h3>What is a circle?</h3>

It is defined as the combination of points that and every point has an equal distance from a fixed point ( called the center of a circle).

We have a hexagon with a side length of 10 feet.

We know the area of the hexagon is given by:

\rm A = \frac{3\sqrt{3} }{2} a^2  where a is the side length.

\rm A = \frac{3\sqrt{3} }{2} 10^2 ⇒ 150\sqrt{3} \rm feet^2

We have the shortest length = x feet and from the figure:

2x = 10

x = 5 feet

The radius of the circle r = longer leg

\rm r = x\sqrt{3} \Rightarrow 5\sqrt{3} feet

The area of the circle a = \pi r^2  ⇒ \pi (5\sqrt{3} )^2 \Rightarrow 75\pi \ \rm feet^2  

The area of the shaded region = A - a

\rm =(150\sqrt{3} \ - 75\pi ) \ feet^2

Thus, the area of the shaded region is  \rm (150\sqrt{3} \ - 75\pi ) \ feet^2

Learn more about circle here:

brainly.com/question/11833983

7 0
2 years ago
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