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Fofino [41]
3 years ago
10

Given that y varies directly to the square root of (x + 1), and that y = 1 when x = 8,

Mathematics
1 answer:
hram777 [196]3 years ago
4 0

Answer:

a) y=1/3 sqrt(x+1)

b) 2/3

c) 224

Step-by-step explanation:

a) y varies directly to sqrt(x+1)

means there is some constant k such that y=ksqrt(x+1)

y=1 when x=8  (plug this in and find k)    1=ksqrt(8+1)

                                                                 1=ksqrt(9)

                                                                 1=k(3)

                                                                 1/3=k

So the equation is y=1/3 * sqrt(x+1)

a)  y=1/3 *sqrt(x+1)

b) what is y when x=3 if you have y=1/3 * sqrt(x+1)

Plug in 3 for x            

y=1/3 *sqrt(3+1)

y=1/3 *sqrt(4)

y=1/3 *2

y=2/3

b) 2/3

c) what is x when y=5

Plug in 5 for y

5=1/3 *sqrt(x+1)

multiply both sides by 3

15=sqrt(x+1)

square both sides

225=x+1

subtract 1 on both sides

224=x

c) 224

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3 0
4 years ago
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The sum of sequences are known as series. Given the following

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