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grigory [225]
3 years ago
10

PLEASE HELP WITH THIS QUESITION WILL GIVE BRAINLIST TO BEST ANSWER

Mathematics
1 answer:
Elena L [17]3 years ago
3 0

Step-by-step explanation:

When dividing two exponents, you can use a formula of:

\frac{x^a}{x^b} = x^{a - b}

So your question would return:

\frac{3^2}{3^6} = 3^{2-6} = 3^{-4}

When an exponent is negative, it is equal to 1 divided by the positive number, so for this question, it would be:

\frac{1}{3^4}

Or in its expanded form:

\frac{1}{81}

Hope this helps!

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I believe its $79. 2.00 times 21 is 42, 1.50 times 34 is 51 and 1.00 times 65 is 65. divide the sum of these three numbers and you get 79.
8 0
3 years ago
Sergey is solving 5x2 + 20x – 7 = 0. Which steps could he use to solve the quadratic equation by completing the square? Select t
olya-2409 [2.1K]

Answer:

Step-by-step explanation:

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\mathrm{For\:}\quad a=5,\:b=20,\:c=-7:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\cdot \:5\left(-7\right)}}{2\cdot \:5}\\x=\frac{-20+\sqrt{20^2-4\cdot \:5\left(-7\right)}}{2\cdot \:5}:\quad \frac{-10+3\sqrt{15}}{5}\\x=\frac{-20-\sqrt{20^2-4\cdot \:5\left(-7\right)}}{2\cdot \:5}:\quad -\frac{10+3\sqrt{15}}{5}\\

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\x=\frac{-10+3\sqrt{15}}{5},\:x=-\frac{10+3\sqrt{15}}{5}

7 0
3 years ago
Read 2 more answers
Can anyone help us on this question
Taya2010 [7]
Area = (1/2) * 4 * 5 = 10 cm²
8 0
3 years ago
Hi. can you help me?
PIT_PIT [208]

Answer:

Sure

Step-by-step explanation:

First you need to ____

Next you need to ____

Last, put nothing

That's what you did, so I helped you out.

Is dis good

7 0
3 years ago
Read 2 more answers
I need help with math.............
telo118 [61]
Greetings!

Solve using the properties of equality:
2(x+2)=4x+1-2x
Distribute the Parenthesis.
2x+4=4x+1-2x
Combine like terms.
2x+4=2x+1
There are no possible solutions.

Hope this helps.
-Benjamin

4 0
3 years ago
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