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Ivenika [448]
2 years ago
15

Common factor of 10 and5

Mathematics
2 answers:
IceJOKER [234]2 years ago
6 0
GCF of 10 and 5 is 5
emmasim [6.3K]2 years ago
3 0
Common factors include: 5 and 1
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Figure G is rotated 90Degrees clockwise about the origin and then reflected over the x-axis, forming figure H. On a coordinate p
asambeis [7]

Answer:

The 1st selection is appropriate.

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2nd: the rotation would need to be 90° CCW

3rd, 4th: rotation or double reflection will give the original orientation. This figure is reflected an odd number of times, so has its orientation reversed.

Hope it helps.. Mark brainliest

8 0
3 years ago
Read 2 more answers
Evaluate this expression. 5^-3
sweet-ann [11.9K]

Answer:

\Huge \boxed{\frac{1}{125}}

Step-by-step explanation:

<h2>Order of operations</h2>

PEMDAS

Parenthesis, exponent, multiply, divide, add, and subtract from left to right.

Do exponent.

\displaystyle \frac{1}{5^3}

\displaystyle 5^3=5\times5\times5=125

\displaystyle \frac{1}{125}=125

\large \boxed{\frac{1}{125}}, which is our answer.

Hope this helps!

4 0
3 years ago
This is super important for my grade guys can you please let me know the answer
nignag [31]

Answer: the last one

Step-by-step explanation:

3 0
3 years ago
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
3 years ago
This is the graph of y= x2 + 4x + 7 what is the range of this function?​
trapecia [35]

Answer:

C

Step-by-step explanation:

4 0
2 years ago
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