All categories

2 years ago

Which of the following are prime factors of both 24 and 42? Select all that apply.

Mathematics

2 answers:

vitfil [10]2 years ago

7 0

Answer:

2 and 3

Step-by-step explanation:

prime factors for 42: 2, 3, and 7

prime factors for 24: 2,2,2 and 3

BlackZzzverrR [31]2 years ago

4 0

Answer:

2 and 3

Step-by-step explanation:

So, 24=2x2x2x3

42=2x3x7

3 and 2

You might be interested in

Solve the system of equations.{y=2x−64x−2y=14{y=2x−64x−2y=14use the substitution method.the solution is (4, 14)(4, 14).the solut

Lilit [14]

I did this test and failed. But when I reviewed it showed me the right answers... so here is the answer: C. There is no solution.

3 0

3 years ago

Read 2 more answers

Multiply the following rational expressions and simplify the result

GarryVolchara [31]

Answer:

Step-by-step explanation:

We have to solve the given expression,

$\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}$

$\frac{9y-33y^2-3y^4}{100-49y^2}.\frac{7y^2+17y+10}{14y^2+28y}$ = $\frac{-y(-9+33y+3y^3)}{100-49y^2}.\frac{7y^2+17y+10}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{7y^2+10y+7y+10}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{y(7y+10)+1(7y+10)}{14y(y+2)}$

= $\frac{-y(-9+33y+3y^3)}{(10-7y)(10+7y)}.\frac{(y+1)(7y+10)}{14y(y+2)}$

= $\frac{-3y(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14y(y+2)}$

= $\frac{-3(-3+11y+y^3)}{(10-7y)}.\frac{(y+1)}{14(y+2)}$

= $\frac{3(3-11y-y^3)(y+1)}{(10-7y)(14(y+2)}$

3 0

3 years ago

How many integers in the set {n ∈ Z | 1 ≤ n ≤ 700} are divisible by 2 or 7?

sukhopar [10]

$\large\begin{array}{l}\\\\ \textsf{This question gives us a set}\\\\ \mathsf{S=\{n \in\mathbb{Z}:~1\le n\le 700\}}\\\\ \mathsf{S=\{1,\,2,\,3,\,\ldots,\,699,\,700\}}\\\\\\ \bullet~~\textsf{Set of integers that are divible by 2 (even integers):}\\\\ \mathsf{A=\{n\in \mathbb{Z}:~n=2k,\,k\in\mathbb{Z}\}}\\\\ \mathsf{A=\{\ldots,\,-4,\,-2,\,0,\,2,\,4,\,\ldots\}}\\\\\\ \bullet~~\textsf{Set of integers that are divible by 7:}\\\\ \mathsf{B=\{n\in \mathbb{Z}:~n=7k,\,k\in\mathbb{Z}\}}\\\\ \mathsf{A=\{\ldots,\,-14,\,-7,\,0,\,7,\,14,\,\ldots\}} \end{array}$

___________

$\large\begin{array}{l}\\\\ \textsf{We want to know how many elements there are in the}\\\textsf{following set:}\\\\ \mathsf{S\cap (A\cup B)=(S\cap A)\cup(S\cap B)\qquad(i)} \end{array}$

____________

$\large\begin{array}{l}\\\\ \bullet~~\mathsf{S\cap A=\{n\in\mathbb{N}:~n=2k~~and~~1\le n\le 700,\,k\in\mathbb{Z}\}}\\\\ \mathsf{S\cap A=\{2,\,4,\,6,\,\ldots,\,698,\,700\}}\\\\ \mathsf{S\cap A=\{1\cdot 2,\,2\cdot 2,\,3\cdot 2,\,\ldots,\,349\cdot 2,\,350\cdot 2\}}\\\\\\ \textsf{So, there are 350 elements in }\mathsf{S\cap A:}\\\\ \mathsf{\#(S\cap A)=350.} \\\\\\ \bullet~~\mathsf{S\cap B=\{n\in\mathbb{N}:~n=7k~~and~~1\le n\le 700,\,k\in\mathbb{Z}\}}\\\\ \mathsf{S\cap B=\{7,\,14,\,21,\,\ldots,\,693,\,700\}}\\\\ \mathsf{S\cap B=\{1\cdot 7,\,2\cdot 7,\,3\cdot 7,\,\ldots,\,99\cdot 7,\,100\cdot 7\}} \\\\\\ \textsf{So, there are 100 elements in }\mathsf{S\cap B:}\\\\ \mathsf{\#(S\cap B)=100.} \end{array}$

____________

$\large\begin{array}{l}\\\\ \textsf{Therefore,}\\\\ \mathsf{\#\big[S\cap (A\cup B)\big]}\\\\ =\mathsf{\#\big[(S\cap A)\cup(S\cap B)\big]}\\\\ =\mathsf{\#(S\cap A)+\#(S\cap B)-\#\big[(S\cap A)\cap(S\cap B)\big]}\\\\ =\mathsf{350+100-50}\\\\ =\mathsf{450-50}\\\\ =\mathsf{400~elements.}\\\\\\ \textsf{There are 400 integers in S that are divisible by 2 or 7.} \end{array}$

If you're having problems understanding the answer, try to see it through your browser: brainly.com/question/2105863

$\large\begin{array}{l}\\\\ \textsf{Any doubts? Please, comment below.}\\\\\\ \textsf{Best wishes! :-)} \end{array}$

Tags: <em>set theory divibilility divisible integers union intersection</em>

3 0

3 years ago

What is 6 percent of 35 dollars

lesya692 [45]

The answer is 2.1 US dollars,

4 0

2 years ago

Read 2 more answers

I’m bad at this so can someone give me the answer please?

Kobotan [32]

5. -6a-48
6. 36x+4
7. 30n+40
8. 1.8m+2

Yess

6 0

2 years ago