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Blizzard [7]
3 years ago
12

Kayden is a stunt driver. One time, during a gig where she escaped from a building about to explode(!), she drove at a constant

speed to get to the safe zone that was 160 meters away. After 3 seconds of driving, she was 85 meters away from the safe zone. Let D represent the distance (in meters) from the safe zone after t seconds. Complete the equation for the relationship between the distance and number of seconds.
Mathematics
1 answer:
krek1111 [17]3 years ago
3 0

d=160-25t

Step-by-step explanation:

Let d be the distance between Kayden and safe zone at any time.

It is given that initially d=160

Lrt her speed be v

Since the speed is constant,the distance covered by her in t seconds=v\times t

So,the distance between her and safe zone after t seconds is 160-vt

It is given that after 3 seconds,d=85

So,160-3v=85

3v=75

v=25

So,d=160-3t

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Read 2 more answers
HELPPPP ME PLZZZZ!!!!!!
mihalych1998 [28]
<h3>Answers:</h3>

14.  Each leg is   21\sqrt{2}  units long.

15 (a). The longer leg is 7\sqrt{3}  units long

15 (b). The hypotenuse is 14 units long

=====================================================

Explanations:

For a 45 degree right triangle, aka 45-45-90 triangle, the hypotenuse is equal to sqrt(2) times the short leg. Algebraically we can say

y = x\sqrt{2} where x is the leg and y is the hypotenuse

Let's solve for x

y = x\sqrt{2}\\\\x\sqrt{2} = y\\\\x = \frac{y}{\sqrt{2}}\\\\x = \frac{y\sqrt{2}}{\sqrt{2}*\sqrt{2}} \ \text{rationalizing denominator}\\\\x = \frac{y\sqrt{2}}{2}\\\\

Now plug in the given hypotenuse y = 42, this leads to,

x = \frac{y\sqrt{2}}{2}\\\\x = \frac{42\sqrt{2}}{2}\\\\x = \frac{42}{2}\sqrt{2}\\\\x = 21\sqrt{2}\\\\

---------------------------

For a 30-60-90 triangle, the hypotenuse is double that of the short leg. So the hypotenuse is 2*7 = 14.

The longer leg is equal to sqrt(3) times the short leg. The longer leg is 7\sqrt{3}

5 0
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