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LuckyWell [14K]
3 years ago
13

The function y=-16x^2+16x+32 represents the height in feet of a firework x seconds after it is launched

Mathematics
1 answer:
Nadya [2.5K]3 years ago
8 0

Answer:

36feet

Step-by-step explanation:

Since we are not asked what to find, we can as well find the maximum height reached by the rocket

At max height, velocity is zero

V = dy/dx = 0

dy/dx = -32x + 16

0 = -32x+16

32x = 16

x = 16/32

x = 0.5s

Get the maximum height

y = -16(0.5)^2+16(0.5) + 32

y = -4+8+32

y = 36feet

Hence the max height reached by the rocket is 36feet

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What is the variable y in w= x y/z?
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Multiply by z to get
wz = xy
divide by x
y = wz/x
8 0
3 years ago
The range of the function f(x) = x + 5 is {7, 9). What is the function's domain?​
Maru [420]

Answer:

(2,4)

Step-by-step explanation:

The range is the y values

y = x+5

Let y=7

7 =x+5

Subtract 5

7-5 x+5-5

2=x

Let y= 9

Subtract 5

9-5 x+5-5

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The domain is (2,4)

8 0
3 years ago
Can someone help me with number 10, 11, 12, 15, and 16 I don’t understand them thanks
yanalaym [24]

10)

10^2-48:6+25*3=100-8+75=167

11)

8(\frac{16}{4} )+2^2-11*3=32+4-33=3

12)

(\frac{6}{3} +4)^2:4*7=(2+4)^2:4*7=6^2:4*7=36:4*7=9*7=63

13) (your answer is incorrect)

5(9-4)^2-3^2=5*5^2-3^2=5*25-9=125-9=116

14) (your answer is incorrect)

5^2-2^2*4^2-12=25-4*16-12=25-64-12=-51

15)

(\frac{50}{5^2} )^2:4=(\frac{50}{25} )^2:4=2^2:4=4:4=1

16)

\frac{24+32+30+28}{2}-expression to represent this situation

\frac{24+32+30+28}{2}=\frac{114}{2}=57

Answer: In 1 group 57 students

/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\

P.S. Hello from Russia :^)

4 0
2 years ago
Help pls
Korvikt [17]

Answer:y74

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The cost per week of running a boarding house is partly constant and partly
patriot [66]

Answer:

the cost of running the boarding

house for 600 students is N61,000

Step-by-step explanation:

Let C represents cost

K1 represents first constant

K2 represents second constant

C= k1+k2n

3500 = k1 + 25 k2............. Eqn(1)

6000= k1 + 50 k2 .............. Eqn(2)

Subtract eqn(1) from eqn(2)

2500= 25k2

K2= 2500/25

K2= 100

To get k1 from eqn(1)

3500 = k1 + 25 k2

Substitute the value of k2

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3500= k1 +2500

K1= 3500- 2500

K1= 1000

The equation connecting them;

C= 1000+ 100n

The cost of running the boarding

house for 600 students is

n= 600

C= 1000+ 100(600)

C= 1000+60000

C= N 61,0000

6 0
2 years ago
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