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Lubov Fominskaja [6]
3 years ago
10

The problem is in the picture

Mathematics
1 answer:
Afina-wow [57]3 years ago
4 0
Angles 1 and 2
Angles 3 and 4
Angles 5 and 6
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A rectangle measures 6 inches by 15 inches. If each dimension of the rectangle is dilated by a scale factor of 1/3 to create a n
Brrunno [24]
I think it is A. I could be wrong
4 0
3 years ago
A rectangular school banner has a length of 44 inches, a perimeter of 156 inches, and an area of 1,496 square inches. the cheerl
STatiana [176]

A rectangular school banner has a length of 44 inches, a perimeter of 156 inches, and an area of 1,496 square inches. the cheerleaders make signs similar to the banner. the length of a sign is 11 inches. First solve the width of the rectangle:

1496 sq in/ 44 = 34 in

So the sign has also a width of 34 in and a length of 11 so the area is 34*11 =374 sq in

<span>The perimeter is (34*2) +(11*2) =90 in</span>

3 0
3 years ago
Simplest form for 16/56
VARVARA [1.3K]
The answer is 2/7 because you're finding the greatest common factor of both numbers. The gcf is 8, so divide both by 8 and you'll get 2/7
4 0
3 years ago
gde Sta manut Stree and Lincoln Set orm a right triangle with a trafic light at each intersection. The distance between the traf
earnstyle [38]

The diagram is of a right triangle with

Hypotenuse = 233

One leg of the triangle = 105

Another Leg = <em>we have to find</em>

<em />

<em />

Pythagorean Theorem can be written as:

\text{Leg}^2+\text{AnotherLeg}^2=\text{Hypotenuse}^2

Thus, we can write:

\begin{gathered} 105^2+\text{AnotherLeg}^2=233^2 \\ \text{AnotherLeg}^2=233^2-105^2 \\ \text{AnotherLeg}^2=43264 \\ \text{AnotherLeg}=\sqrt[]{43264} \\ \text{AnotherLeg}=208\text{ ft} \end{gathered}

Thus,

The distance between the traffic lights on Walnut Street is:

208 feet
4 0
1 year ago
What is 1/2 as a decimal?
frosja888 [35]
0.5 just look it up
3 0
3 years ago
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