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Liono4ka [1.6K]
2 years ago
13

Help me please number 3

Mathematics
1 answer:
asambeis [7]2 years ago
3 0

Answer:

c = 26

Step-by-step explanation:

This is a right triangle, so we can use the Pythagorean theorem.

a^2 + b^2 = c^2

where a and b are the legs and c is the hypotenuse

10^2 + 24^2 = c^2

100 + 576 = c^2

676 = c^2

Take the square root of each side

sqrt(676) = sqrt(c^2)

26 = c

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Plssssss help fast!!!!Launching From a Tower
aleksandrvk [35]

Answer:

A. h = 64

B.  t = 4 -> 4 seconds

C. t = 1.5 -> 1.5 seconds

D. maximum height is 100 ft

E. The domain that makes sense for the function in this context is t

is any positive real number since time can not be negative.

Step-by-step explanation:

h = -16t2 + vt + 64

A. What tower platform height was the projectile launched from?

when the projectile was not launched, t = 0

h = -16(0)^2 + v(0) + 64 = 64

B. How long was the projectile in the air?

if the projectile lands, its height = 0 so substitute 0 for h

0 = -16(t)^2 + 48(t) + 64

  = -16(t^2 - 3t -  4)

 = -16(t - 4)(t + 1)

t = 4 or t = -1

Since time can not be negative, t = -1 can not be the answer. Therefore, the projectile lands when t = 4 or 4 seconds.

C. When did it reach its maximum height?

maximum -> t=-b/2a where in -16(t)^2 + 48(t) + 64, b = 48 and a = -16

t = -48/-32 = 1.5

D. What was its maximum height?

plug t = 1.5 into -16(t)^2 + 48(t) + 64

-16(1.5)^2 + 48(1.5) + 64 = -36 + 72 + 64 = 100

E. The domain that makes sense for the function in this context is t

is any positive real number since time can not be negative.

6 0
2 years ago
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