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3 years ago

Lots of points:

2 answers:

5 0

480 + 192 = 672 students total
16 + 12 = 28 classrooms total
672/28 = 24 students per room
24 * 12 = 288 students that need to be in school B
288 - 192 = 96 students who have to transfer

Marta_Voda [28]3 years ago

3 0

School B is the right answer

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800 is 100 times as much as

Slav-nsk [51]

800/100 = 8
800 is 100 times as much as 8

8 0

3 years ago

If a graph of y=-4x +2 were changed to a graph of y=-4x+5, how would the y- intercept change?

Dmitry [639]

Answer:

So if y=-4x+2 was changed to y=-4x+5, then the y-intercept would increase by 3.

The y-intercept was (0,2) then it becomes (0,5) in the new line.

Step-by-step explanation:

The slope-intercept form a line is y=mx+b where m is the slope and the y-intercept is b.

Both of these equations given are in this form.

y=-4x+2 when compared to y=mx+b you see that m=-4 and b=2.

Since b=2 then the y-intercept is 2.

y=-4x+5 when compared to y=mx+b you see that m=-4 and b=5.

Since b=5 then the y-intercept is 5.

So if y=-4x+2 was changed to y=-4x+5, then the y-intercept would increase by 3.

8 0

3 years ago

Find each of the numbers below.

iren [92.7K]

Answer:

opposite of -7 is 7 the oppositeof 3 is -3 oppositeof -4 is 4

8 0

3 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .