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3 years ago

Can anyone help me with these three problems?

Mathematics

1 answer:

Galina-37 [17]3 years ago

3 0

Answer:

the first one is 3.31

Step-by-step explanation:

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Tanya [424]

Answer:

$2.50$

Step-by-step explanation:

Given

$(x_1,y_1) = (-1,0.8)$

$(x_2,y_2) = (0,2)$

$(x_3,y_3) = (1,5)$

$(x_4,y_4) = (2,12.5)$

Required

The rate of change (b)

The above graph is represented as:

$y = ab^x$

For: $(x_2,y_2) = (0,2)$ ;

We have:

$y = ab^x$

$2 = a * b^0$

$2 = a *1$

$2 = a$

$a = 2$

For $(x_3,y_3) = (1,5)$ ,

We have:

$y = ab^x$

$5 = a * b^1$

$5 = a * b$

Substitute $a = 2$

$5 = 2 * b$

Divide by 2

$2.5 = b$

$b = 2.5$

<em>Hence, the rate of change is 2.50</em>

7 0

3 years ago

What is the value of c so that -12 and 12 are both solutions of x^2-c=108?

murzikaleks [220]

Answer:

c=36

Step-by-step explanation:

Because $x^2-c=108$ can be rewritten as $x^2-c-108=0$ , finding the value of c so that x=(+-)12 solves this equation it means to find the value of c so that $(+-)12^2-c-108=0$ .
Because $12^2=(-12)^2=144$ , we must find c so that $144-c-108=0$ .
Because $144-108=36$ , c=36.
To check, replace c=36 in the initial equation, and check it is a solution for x=(+-)12.