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3 years ago

Help would be very much appreciated I do not understand. Thanks.

Mathematics

1 answer:

vagabundo [1.1K]3 years ago

7 0

Answer:

im pretty sure it is 2 and 8

Step-by-step explanation:

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4 years ago

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3 years ago

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3 years ago

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sergiy2304 [10]

We have been given a graph of function g(x) which is a transformation of the function $f(x)=3^x$

Now we have to find the equation of g(x)

Usually transformation involves shifting or stretching so we can use the graph to identify the transformation.

First you should check the graph of $f(x)=3^x$

You will notice that it is always above x-axis (equation is x=0). Because x-axis acts as horizontal asymptote.

Now the given graph has asymptote at x=-2

which is just 2 unit down from the original asymptote x=0

so that means we need shift f(x), 2 unit down hence we get:

$y=3^x$

but that will disturb the y-intercept (0,1)

if we multiply $3^x$ by 3 again then the y-intercept will remain (0,1)

Hence final equation for g(x) will be:

$g(x)=3(3^x)-2$

5 0

3 years ago

If 10000 is invested at an interest rate of 10 per year ,compound semiannually,find the value of the investment after the given

slega [8]

Answer:

Part a) $\$17,958.56$

Part b) $\$32,251.00$

Part c) $\$57,918.16$

Step-by-step explanation:

<u><em>The complete question is</em></u>

If $10,000 is invested at an interest rate of 10% per year, compounded semiannually, find the value of the investment after the given number of years. (Round your answers to the nearest cent.)

a)6 years

b)12 years

c)18 years

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part a) 6 years

we have

$t=6\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2$

substitute in the formula above

$A=10,000(1+\frac{0.10}{2})^{2*6}$

$A=10,000(1.05)^{12}$

$A=\$17,958.56$

Part b) 12 years

we have

$t=12\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2$

substitute in the formula above

$A=10,000(1+\frac{0.10}{2})^{2*12}$

$A=10,000(1.05)^{24}$

$A=\$32,251.00$

Part c) 18 years

we have

$t=18\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2$

substitute in the formula above

$A=10,000(1+\frac{0.10}{2})^{2*18}$

$A=10,000(1.05)^{36}$

$A=\$57,918.16$

7 0

4 years ago