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Margarita [4]
3 years ago
11

Evaluate negative 3 to the power of 2 + (2-6)(10) Please help!

Mathematics
1 answer:
Tom [10]3 years ago
7 0

Answer:

(2-6)(10) plus 3 to the power of 2, is <u>-31</u>.

Explanation:

Three to the power of two plus parenthesis two minus six parenthesis parenthesis ten. This is the same as the sum or addition of 3 times 3(3 times itself two times), and 2 minus 6 times 10.

This expression can be represented by (3^2) + (2-6)(10) which is (3 × 3) + (-4 × 10) which is 9 + -40 which simplifies to -31

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The box method or latitude or longitude.
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2 years ago
Solve for X, 132 is an exterior angle. Show work please!
Darina [25.2K]

Answer:

105/9

Step-by-step explanation:

180 - 132 = 48 degrees

180 - 48 - 45

= 180 - 93

= 87 degrees

<2 = 180 - 87

= 93 degrees

=> 9x - 12 = 93

=> 9x = 105

=> x = 105/9

6 0
2 years ago
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Greeley [361]

\large\displaystyle\text{$\begin{gathered}\sf \frac{2x-20}{3}=2x  \end{gathered}$}

Multiply both sides of the equation by 3.

\large\displaystyle\text{$\begin{gathered}\sf 2x-20=6x \end{gathered}$}

Subtract 6x on both sides.

\large\displaystyle\text{$\begin{gathered}\sf 2x-20-6x=0 \end{gathered}$}

Combine 2x and −6x to get −4x.

\large\displaystyle\text{$\begin{gathered}\sf -4x-20=0 \end{gathered}$}

Add 20 to both sides. Any value plus zero results in its same value.

\large\displaystyle\text{$\begin{gathered}\sf -4x=20 \end{gathered}$}

Divide both sides by −4.

\large\displaystyle\text{$\begin{gathered}\sf x=\frac{20}{-4}  \end{gathered}$}

Divide 20 by −4 to get −5.

\large\displaystyle\text{$\begin{gathered}\sf x=-5 \ \ \to \ \ \ Answer \end{gathered}$}

<h2>{ Pisces04 }</h2>
5 0
1 year ago
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every prime number greater than 10 has a digit in the ones place that is included in which set of numbers below?
stich3 [128]
So we can start with the full of possibilities and eliminate them one by one.

The full set is {0,1,2,3,4,5,6,7,8,9}.

Now we know that any prime greater than 2 is odd as otherwise it would have 2 as a factor, so we can eliminate all of these digits that would be an even number, leaving:

{1,3,5,7,9}

We also know that any prime greater than 5 cannot be a multiple of 5 and that all numbers with 5 in the digits are a multiple of 5, so we can eliminate 5.

{1,3,7,9}

We know that 11,13,17 and 19 are all primes, so we cannot eliminate any more of these, leaving the set:

{1,3,7,9} as our answer.
8 0
3 years ago
Factor by using the perfect-square trinomial formula. <br><br> 100x^2+20x+1
a_sh-v [17]
To factor first multily 100 and 1 and get 100

then find what 2 numbers multiply to get 100 and add to get 20

the numbers are 10 and 10

so
split the center term up
100x^2+10x+10x+1
group
(100x^2+10x)+(10x+1)
undistribute
(10x)(10x+1)+(1)(10x+1)
undistribute/reverse distributive property
(10x+1)(10x+1)
(10x+1)^2
8 0
3 years ago
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