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UNO [17]
3 years ago
7

The function f(x) =2(3)^x is to be multiplied by the function g(x) 3(3)^(2x) to create the function h(x). Which function is prod

uced?
A. h(x)=6(9)^(3x)

B. h(x)=6(3)^(2x^2)

C. h(x)=6(3)^(3x)

D. h(x)=6(6)^(2x^2)
Please show work!
Mathematics
1 answer:
maria [59]3 years ago
7 0

Answer C is the right answer.

Step-by-step explanation:

h(x) = f(x) * g(x)

h(x) = (2*3^x ) * ( 3*3^2x )

2 * 3 is easy, that will be 6.

The ground number 3 remains 3 in h(x), so that is easy too...

But with multiplying exponents, you can add them.

Let's concentrate only on the exponents of f(x) and g(x)... and add them...

x + 2x =3x

So, now combine the easy part with this new exponent, and you get <u>h(x)</u><u> </u><u>=</u><u> </u><u>6</u><u>*</u><u>(3)^(3x)</u>

<u>So</u><u> </u><u>answer C is the right answer.</u>

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Step-by-step explanation:

1.Add all the numbers up which is 935.

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Alexus [3.1K]

Answer:

a) the common difference is 20

b) x_8=115 , x_{12}=195

c) the common difference is -13

d) a_{12}=52, a_{15}=13

Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

36-16 =20

So, the common difference is 20

b) what is x_8? what is x_12

The formula used is: x_n=x_1+(n-1)d

We know common difference d= 20, we need to find x_1

Using x_3=16 we can find x_1

x_n=x_1+(n-1)d\\x_3=x_1+(3-1)d\\15=x_1+2(20)\\15=x_1+40\\x_1=15-40\\x_1=-25

So, We have x_1 = -25

Now finding x_8

x_n=x_1+(n-1)d\\x_8=x_1+(8-1)d\\x_8=-25+7(20)\\x_8=-25+140\\x_8=115

So, \mathbf{x_8=115}

Now finding x_{12}

x_n=x_1+(n-1)d\\x_{12}=x_1+(12-1)d\\x_{12}=-25+11(20)\\x_{12}=-25+220\\x_{12}=195

So, \mathbf{x_{12}=195}

c) what is the common difference of the sequence a_m

Looking at the table, we get a_7=104, a_8=91 and a_9= 78

Deterring the common difference by subtracting a_7 from a_8 we get

91-104 =-13

So, the common difference is -13

d) what is a_12? what is a_15?

The formula used is: a_n=a_1+(n-1)d

We know common difference d= -13, we need to find a_1

Using a_7=104 we can find x_1

a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195

So, We have a_1 = 195

Now finding a_{12} , put n=12

a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52

So, \mathbf{a_{12}=52}

Now finding a_{15} , put n=15

a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13

So, \mathbf{a_{15}=13}

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