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3241004551 [841]
3 years ago
8

Solve for the variable please

Mathematics
2 answers:
spayn [35]3 years ago
8 0

Answer:

n = 0

Step-by-step explanation:

Anything to the power of 0 is 1

x^12 is already equal to x^12 so it only needs to be multiplied by 1

So n = 0

labwork [276]3 years ago
7 0

Answer:

Step-by-step explanation:

x^{12}*x^{n}=x^{12}\\\\x^{12+n}=x^{12}

Compare the powers

12 + n =12

n = 12-2 = 0

x^{n} = x^{0} = 1\\

Anything to the power of zero is 1

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Please help thank you
Dvinal [7]

the tale-tell fellow is the number inside the parentheses.

if that number, the so-called "growth or decay factor", is less than 1, then is Decay, if it's more than 1, is Growth.

\bf f(x)=0.001(1.77)^x\qquad \leftarrow \qquad \textit{1.77 is greater than 1, Growth} \\\\[-0.35em] ~\dotfill\\\\ f(x)=2(1.5)^{\frac{x}{2}}\qquad \leftarrow \qquad \textit{1.5 is greater than 1, Growth} \\\\[-0.35em] ~\dotfill\\\\ f(x)=5(0.5)^{-x}\implies f(x)=5\left( \cfrac{05}{10} \right)^{-x}\implies f(x)=5\left( \cfrac{1}{2} \right)^{-x} \\\\\\ f(x)=5\left( \cfrac{2}{1} \right)^{x}\implies f(x)=5(2)^x\qquad \leftarrow \qquad \textit{Growth} \\\\[-0.35em] ~\dotfill

\bf f(t)=5e^{-t}\implies f(t)=5\left( \cfrac{e}{1} \right)^{-t}\implies f(t)=5\left( \cfrac{1}{e} \right)^t \\\\\\ \cfrac{1}{e}\qquad \leftarrow \qquad \textit{that's a fraction less than 1, Decay}

now, let's take a peek at the second set.

\bf f(x)=3(1.7)^{x-2}\qquad \leftarrow \qquad \begin{array}{llll} \textit{the x-2 is simply a horizontal shift}\\\\ \textit{1.7 is more than 1, Growth} \end{array} \\\\[-0.35em] ~\dotfill\\\\ f(x)=3(1.7)^{-2x}\implies f(x)=3\left(\cfrac{17}{10}\right)^{-2x}\implies f(x)=3\left(\cfrac{10}{17}\right)^{2x} \\\\\\ \textit{that fraction is less than 1, Decay} \\\\[-0.35em] ~\dotfill

\bf f(x)=3^5\left( \cfrac{1}{3} \right)^x\qquad \leftarrow \qquad \textit{that fraction is less than 1, Decay} \\\\[-0.35em] ~\dotfill\\\\ f(x)=3^5(2)^{-x}\implies f(x)=3^5\left( \cfrac{2}{1} \right)^{-x}\implies f(x)=3^5\left( \cfrac{1}{2} \right)^x \\\\\\ \textit{that fraction in the parentheses is less than 1, Decay}

6 0
4 years ago
**100 POINTS**I WILL MARK AS BRAINLIEST IF YOU GET IT CORRECT!
Akimi4 [234]

Answer:

As we know:

Direct variation => y = k*x, k is constant ( y/x = k = constant)

Inverse variation => y = k/x, k is constant (y*x = k = constant)

X - 1 2 5 10

Y - 6 12 30 60

Check the first 2 pair of X, Y:

(the remaining pair of value would show the same tendency)

X1 x Y1 = 1 x 6 = 6

X2 x Y2 = 2 x 12 = 24

=> X x Y is not a constant

=> Not an inverse variation.

=> B and D is incorrect

Otherwise, we have:

Y1/X1 = 6/1 = 6

Y2/X2 = 12/2 = 6

=> Y/X = 6

=> A direct variation with k = 6

=> y = 6x

=> Option C is correct

Hope this helps!

:)

6 0
3 years ago
Read 2 more answers
PLEASE HELP 10 POINTS
abruzzese [7]
A) 70 pools in total
B) 5:2
5 0
3 years ago
F(x)= 2x^2 + k/x point of inflection at x=-1 then value of k is
Nady [450]
I think it isss 2 :)))
4 0
3 years ago
¿Can i get the answer?
NikAS [45]

Answer:

15.7

Step-by-step explanation:

5 x 3.14(pi) = 15.7

3 0
3 years ago
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