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Lina20 [59]
3 years ago
9

Henry and Blake are both runners. In 2 hours, Henry runs 14 miles. In 5 hours, Henry runs 35 miles. The graph below represents t

he number of miles Blake runs in x hours.
A. Blake runs at a slower speed.
B. There is not enough information to determine their speed.
C. Henry and Blake run at the same speed.
D. Henry runs at a slower speed.
Mathematics
1 answer:
Tasya [4]3 years ago
3 0
Henry runs at a rate of seven miles an hour. I think you meant the second sentence to be 'Blake', because there is not attatched graph. They are both running at the same speed in this case.

I got this because 14 (miles ran by Henry) / 2 (hours ran) you get 7. This same equation is applied to (Blake?) 35 (miles ran by [Blake?]) / 5 (hours ran) also equals seven.

If you meant the second statement to be Blake, they are both running at the same speed, and x=7.

If you didn't, then there is not enough information to determine Blake's speed.

If you found this especially helpful, I'd appreciate if you'd vote me Brainliest for your answer. I want to be able to assist more users one-on-one, as well as to move up in rank! :)
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3 years ago
Please find the perimeter and area of these shapes
ladessa [460]

Answer:

ABC shaded area = 36\pi - 72   cm²

ABC shaded area perimeter = 6\pi +12\sqrt{2}    cm

ABCD area = \dfrac52 \pi  cm²

ABCD perimeter = 3\pi +2   cm

Step-by-step explanation:

<u>Shape ABC</u>

Assuming you want the area and perimeter of the shaded part of the shape only...

<u>Area</u>

Area of a sector = \dfrac12r^2\theta (where r is the radius and \theta<em> </em>

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⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²

Therefore, area of shaded area = area of sector - area of triangle

⇒ area = 36\pi - 72 cm²

<u>Perimeter</u>

Arc length = r\theta (where r is the radius and \theta<em> </em>

⇒ arc length = 12\times\dfrac12\pi =6\pi  \ \textsf{cm}

Hypotenuse of triangle = \sqrt{a^2+b^2} (where a and b are the legs of the right triangle)

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⇒ area of large semicircle ABC = \dfrac12 \times \pi \times 2^2=2\pi  \ \textsf{cm}^2

⇒ area of small semicircle AD = \dfrac12 \times \pi \times 1^2=\dfrac12\pi  \ \textsf{cm}^2

⇒ area of shape ABCD = \dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2

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1/2 circumference = \pi r

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Answer:

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The line crosses the y- axis at (0, 5 ) ⇒ c = 5

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