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tresset_1 [31]
3 years ago
11

How could you test if a number is divisible by 12 15 or 24

Mathematics
1 answer:
Rina8888 [55]3 years ago
3 0

Answer:

you will find out if the number is a multiple of the numbers.

Step-by-step explanation:

3×5=15

3×4=12

3×8=24

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72 = –6(y + 37)<br> plz help
AlladinOne [14]

Answer:

-49

Step-by-step explanation:

you can first divide both sides by -6

so

-12 = y+37

so y = -12-37 = -49

8 0
2 years ago
Select all of the following that are quadratic equations.​
Nesterboy [21]

Answer:

x^2 -2x = 4x+1

2x^2 +12x = 0

9x^2 +6x -3=0

Step-by-step explanation:

A quadratic equation has the highest power of x being squared

x^2 -2x = 4x+1

2x^2 +12x = 0

9x^2 +6x -3=0

These are all quadratic equations

5 0
3 years ago
Read 2 more answers
Which statement is true about the angles?
liq [111]

Complete Question:

Triangle abc has the angle mesausres shown.

m<A = (2x)°

m<B = (5x)°

m<C = (11x)°

Which statement is true about the angles?

A. m<A = 20°

B. m<B = 60°

C. m<A and m<B are complementary

D. m<A + m<C = 120°

Answer:

A. m<A = 20°

Step-by-step explanation:

m<A + m<B + m<C = 180° (sum of interior angles of a triangle)

2x + 5x + 11x = 180 (substitution)

Solve for x. Add like terms.

18x = 180

Divide both sides by 18

\frac{18x}{18} = \frac{180}{18}

x = 10

Find the measure of each angle by substituting x = 10:

m<A = (2x)° = 2(10) = 20°

m<B = (5x)° = 5(10) = 50°

m<C = (11x)° = 11(10) = 110°

Therefore, the only true statement is:

A. m<A = 20°

3 0
2 years ago
Read 2 more answers
A line is defined by the equation y = two-thirds x minus 6. The line passes through a point whose y-coordinate is 0. What is the
DedPeter [7]

y = \cfrac{2}{3}x~~ - ~~6~\hspace{10em} (\stackrel{x}{?}~~,~~\stackrel{y}{0}) \\\\\\ 0=\cfrac{2}{3}x~~ - ~~6\implies 6=\cfrac{2x}{3}\implies 18=2x\implies \cfrac{18}{2}=x\implies 9=x

3 0
2 years ago
For which pair of functions is (gºf)(a) =la|-2
nikklg [1K]

Answer:

<h2>              The third</h2>

Step-by-step explanation:

(g\circ f)(x) = g\big(f(x)\big)

so:

I.\quad (g\circ f)(a)=\sqrt{f(a)}=\sqrt{a^2-4}=\sqrt{(a+2)(a-2)}\\\\II.\quad (g\circ f)(a)=2f(a)-2=2(\frac12a-1)-2=a-2-2=a-4\\\\ \bold{III.\quad (g\circ f)(a)=\sqrt{f(a)-5}-2=\sqrt{5+a^2-5}-2=\sqrt{a^2}-2=|a|-2}\\\\ IV.\quad (g\circ f)(a)=4f(a)-5=4(3-3a)-5 = 12-12a-5=-12a+7

8 0
3 years ago
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