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2 years ago

What is the answer to (12^2)^3 times 12^-4 and 12^7 times 12^-5 over 12^2

Mathematics

2 answers:

raketka [301]2 years ago

7 0

Answer: 12² = 144

12² ÷ 12² = 1

Step-by-step explanation:

I'll go through it backwards: 12^7 times 12^-5 You subtract the -5 from 7 in the exponents to get 12^2. 12^2 divided by 12^2 is 1.

(12^2)^3 works out to 12^6 because you multiply the exponents when its a power raised by a power. Then you subtract the 4 from the 6 (or add -4 +6) to get 12^2.

That means 12 squared, same as 12 times 12

That's 144.

If you have to multiply the two parts together, It's 144 times 1.

If "and" means "add them together"-- YOU KNOW!

shtirl [24]2 years ago

6 0

Answer:

Step-by-step explanation:

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Anton [14]

Answer:

$\sqrt{44} = 2\sqrt{11}$

Step-by-step explanation:

Apply pythagorean theorem:

ABD is a right triangle with angle ADB = 90

hence AD = $\sqrt{AB^{2} - BD^{2} }$ where BD = 10 because it is half of BC

AD = $\sqrt{12^{2}-10^{2} }$ = $\sqrt{44} = 2\sqrt{11}$

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2 years ago

Solve for the variable -3√(t) = -21

fenix001 [56]

Answer:

Step-by-step explanation:

-3√(t) = -21

√(t) = 7

t = 49

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2 years ago

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Natalka [10]

Answer:

1,rational

2,irrational

3,rational

4,rational

5,irational

Step-by-step explanation:

for number 1 and 2just ask your self what times what =this and

number 4 just divide 72 by 4 and

number 5 pie is always irrational

6 0

2 years ago

All functions have a domain of R (all real numbers). True Or False

Veronika [31]

Answer:

True

Step-by-step explanation: Trust me

4 0

2 years ago

0 write the quation of the graph

Vika [28.1K]

to get the equation of any straight line, all we need is two points from it, Check the picture below, let's use those two.

$(\stackrel{x_1}{-5}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-5)}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-5)}}}\implies \cfrac{1+5}{0+5}\implies \cfrac{6}{5}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-5)}=\stackrel{m}{\cfrac{6}{5}}(x-\stackrel{x_1}{(-5)}) \\\\\\ y+5=\cfrac{6}{5}(x+5)\implies y+5=\cfrac{6}{5}x+6\implies y=\cfrac{6}{5}x+1$

8 0

1 year ago