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Stells [14]
2 years ago
14

Find the area of the shape above

Mathematics
1 answer:
kirill115 [55]2 years ago
6 0

Answer:

625 cm^2

Step-by-step explanation:

Break it into a rectangle on the bottom and a triangle on top

rectangle area = 50 x 8 = 400 cm^2

triangle area = 1/2 b * h = 1/2 * 50 * 9 = 225 cm^2

added together = 625 cm^2

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Pls help me and give steps as to how to do it thank you
andrew11 [14]

Answer: x=13, y=132

Step-by-step explanation:

Using the same-side interior angles theorem, y=132.

Using the alternate interior angles theorem, 5x-17=48 \implies x=13

7 0
1 year ago
Find the radius of a circle if the circumference is 66cm​
FrozenT [24]

Answer:

Approximately 10.5, or 10.5095541.

Step-by-step explanation:

To find the radius given the circumference of a circle, we first have to divide the circumference by pi, or an estimation of 3.14. This will give us the diameter.

Since we know that the radius of a circle is simply half the diameter, we divide the number we got from the problem above by 2. This will give you your answer!

7 0
3 years ago
Read 2 more answers
Can someone help me with this two step equation
hjlf
You have to distribute the 4 to the numbers in the parenthesese. So 4 times 3x which is 12x and 4 times 5 which is 20 so now you get 12x plus 20 equals 8. Now you have to subtract 20 from 8 which is -12 so now you have 12x=-12 now divide the 12 from x and 12 from -12 and so now you have x=-1
3 0
3 years ago
Given the equation, =+ solve for t
Olenka [21]

Answer:

I don't see an equation or anything???

Step-by-step explanation:

4 0
3 years ago
A basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modele
aev [14]

THIS IS THE COMPLETE QUESTION

basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modeled by the equation h = 6 + 30t – 16t2. Two-tenths of a second after the shot is launched, an opposing player leaps up to block the shot. The height of the shot blocker’s outstretched hands t seconds after he leaps is modeled by the equation h = 9 + 25t – 16t2. If the ball reaches the net 1.7 seconds after the shooter launches it, does the leaping player block the shot?

Yes, exactly 0.6 seconds after the shot is launched.

Yes, between 0.64 seconds and 0.65 seconds after the shot is launched

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

No, the shot is not blocked.

Answer:

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

Given that the

equation for shot block height h = 9 + 25t – 16t2.

equation for ball height h = 6 + 30t – 16t2.

From the question ,it can be deducted that the shot is made before two tenths of a second or 0.2seconds

CHECK THE ATTACHMENT TO COMPLETE THE DETAILED SOLUTION

5 0
3 years ago
Read 2 more answers
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