All categories

4 years ago

Which of these numbers is divisible by both 2 and 5? A) 552 B) 255 C) 525 D) 250

Mathematics

1 answer:

artcher [175]4 years ago

7 0

Answer:

D) 250

Step-by-step explanation:

A) 552

552 ÷ 2 = 276

552 ÷ 5 = 110.4

B) 255

255 ÷ 2 = 127.5

255 ÷ 5 = 51

C) 525

525 ÷ 2 = 262.5

525 ÷ 5 = 105

D) 250

250 ÷ 2 = 125

250 ÷ 5 = 50

D is the only answer where both 2 and 5 divide evenly, so it is your answer.

Hope this helps :)

You might be interested in

The likelihood that Sue makes a free throw in basketball is 73%. The likelihood that she makes a 3-point shot is 0.72. Which is

AleksandrR [38]

Answer:

She is more likely to make a free throw in basketball.

Step-by-step explanation:

This is because 73% is equal to .73 and .72 is equal to 72% so thats more common.

5 0

4 years ago

What is the slope-intercept form in the picture

grandymaker [24]

Answer:

y= 3/2x+40

Step-by-step explanation:

y= rise/run+ y-intercept

5 0

3 years ago

(3)-2<br> Evaluate exponent

olga nikolaevna [1]

Answer:

$3^{-2} = \dfrac{1}{9}$

Step-by-step explanation:

Definition of negative exponent:

$a^{-n} = \dfrac{1}{a^n}$

In our case, a = 3 and -n = -2.

$3^{-2} = \dfrac{1}{3^2}$

Now we just evaluate the power in the denominator.

$3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$

7 0

3 years ago

Answer this please !

kompoz [17]

Answer:

x = 90°

Step-by-step explanation:

The diagonals of the kite AC and BD are perpendicular to each other

Hence x = 90°

6 0

3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago