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Lesechka [4]
3 years ago
9

Simplify the expression x^5 • x^7

Mathematics
2 answers:
sweet [91]3 years ago
8 0

Answer:

x^12

Step-by-step explanation:

just add the exponents when multiplying

iragen [17]3 years ago
7 0

Answer:

\Huge\boxed{\mathsf{\Rightarrow X^1^2}}

Step-by-step explanation:

<h3><u>TO SOLVE:</u></h3>

With exponent rule.

<h3><u>SOLUTIONS:</u></h3>

First, use exponent rule.

<u><em>EXPONENT RULE:</em></u>

<u><em /></u>

\displaystyle \mathsf{\Rightarrow A^B*A^C=A^B^+^C}}

\displaystyle \mathsf{X^5^+^7}}

Solve.

Add the numbers from left to right.

\displaystyle \mathsf{7+5=\boxed{\mathsf{\Rightarrow 12}}}}

\Rightarrow \Large \boxed{\mathsf{X^1^2}}}

So, the final answer is x¹².

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50 POINTS!!
Charra [1.4K]

Answer:

FH=73

Step-by-step explanation:

We are given that a segment FH and G is the midpoint of FH.

FH=14x-4

FG=5x+9

We have to find the measure of FH.

When G is the midpoint of segment FH

Then , FG=GH

FH=FG+GH

Segment addition property

14x-4=FG+FG

Substitution property

14x-4=2FG

14x-4=2(5x+9)

14x-4=10x+18

14x-10x=18+4

4x=22

x=\frac{22}{4}=5.5

Substitute the values then we get

FG=5(5.5)+9=36.5

FG=GH=36.5

FH=14(5.5)-4=73

8 0
3 years ago
If 0 &lt; x &lt; 1 and 0 &lt; y &lt; 1, which of the following must be true?
katovenus [111]
0<x<1 and 0<y<1

x>0 so x is positive and y>0 so y is also positive.
When you multiply two positive numbers you always get a positive number, so the product of x and y must be positive, or greater than 0.
xy>0 - it must be true
xy<0 - it can't be true
Also when you divide a positive number by a positive number you always get a positive number, so the quotient of x and y must be positive.
x/y<0 - it can't be true
D and E can be true, but don't have to. It depends on the values of x and y. If x>y, then x-y>0 is true and x-y<0 isn't true; if x<y, then x-y>0 isn't true and x-y<0 is true.

Therefore, only A <u>must</u> be true.
3 0
4 years ago
PLZ HELP!!!!!!! (20 points)
AlekseyPX

Answer:

Neither A nor B

Step-by-step explanation:

The set of pairs, where the front member of every sequence pair must be unique.

3 0
3 years ago
Please help, whoever gets it right I will mark brainliest.
Pepsi [2]

Answer:

One and Four

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
I need help simplifying a math problem. It is (x^2 * x^-3)^4
Mashutka [201]
(x^2 \times x^{-3})^4 \\ \\ (x^{2-3})^4 \ / \ product \ rule \\ \\ (x^{-1})^4 \ / \ simplify \\ \\ x^{-4} \ / \ simplify \\ \\ \frac{1}{x^4} \ / \ negative \ power \ rule \\ \\ Answer: \frac{1}{x^4}
4 0
4 years ago
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