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3 years ago

Apply the distributive property and the greatest common factor to write an equivalent expression.

Mathematics

2 answers:

shepuryov [24]3 years ago

8 0

Answer:

Equivalent expression of 60x−24 in distributive property is

12 × (5x - 2) = 12 ×5x - 12 × 2 .

Step-by-step explanation:

Distributive property

Let us assume that the a, b, c are any number .

than a (b+c) = a.b + a.c

This property is called distributive property .

As given the expression

= 60x - 24

It is written as in the distributive property form .

12 × (5x - 2) = 12 ×5x - 12 × 2

Therefore the equivalent expression of 60x−24 in distributive property is

12 × (5x - 2) = 12 ×5x - 12 × 2 .

myrzilka [38]3 years ago

3 0

You can take out a 12 from each factor as they can both be divided by 12. That leaves you with 12(5x-2) if you applied the distribution property, you'd end up with the same answer you started with.

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What’s the value of x? :/

kolbaska11 [484]

Answer:

$x = 27$

Step-by-step explanation:

According to congruent chords and arcs theorem, if $\overline{XY} \cong \overline{ZY}$ then $\arc{XY} \cong \arc{ZY}$ .

This means that $\arc{XY} = \arc{ZY} = (6x - 20) degrees$

Thus:

$\arc{XY} + \arc{ZY} + \arc{ZX} = 360$ (full circle = 360°)

$\arc{XY} = (6x - 20)$

$\arc{ZY} = (6x - 20)$

$\arc{ZX} = 76$

Plug in the values into the equation

$(6x - 20) + (6x - 20) + 76 = 360$

$6x - 20 + 6x - 20 + 76 = 360$

Add like terms

$12x + 36 = 360$

Subtract 36 on both sides

$12x = 360 - 36$

$12x = 324$

Divide both sides by 12

$x = 27$

8 0

3 years ago

Write in y=mx+b form

DENIUS [597]

Step-by-step explanation:

b =-4 (y-intercept)

x =0

m = 0 (the line doesn't have a slope)

y = mx + c

y = 0(0) + (-4)

y = - 4

7 0

3 years ago

Does (-2)^n/( 3^n+1) converge or diverge?

hodyreva [135]

$\left|\dfrac{(-2)^n}{3^n+1}\right|=\dfrac{2^n}{3^n+1}$

As $n\to\infty$ , the sequence $a_n=\left(\dfrac23\right)^n$ converges to zero.

If you're talking about the infinite series

$\displaystyle\sum_{n\ge0}\dfrac{(-2)^n}{3^n+1}$

well we've shown by comparison that this series must also converge because we know any geometric series $\sum\limits_n r^n$ will converge as long as $|r|$ .