All categories

2 years ago

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

You might be interested in

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.

_____

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:

Step-by-step explanation:

4 0

2 years ago