Help don't know how to do question

Mathematics

1 answer:

Gre4nikov [31]3 years ago

4 0

You have to multiple b× w×h. the answer would be 342

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Help with these 2 please !!

stiks02 [169]

The answer is A and B

3 0

3 years ago

Given a rectangle with side lengths of and , what is its area?

maw [93]

B = 2x+4
H = 3x-5

A = bh
= (2x+4)(3x-5)
= (2x)(3x)+(2x)(-5)+(4)(3x)+(4)(-5)
= 6x^2 - 10x + 12x - 20
= 6x^2 + 2x - 20

So the answer is A (6x^2 + 2x - 20) (i think)

8 0

3 years ago

Read 2 more answers

Triangle pqr has vertices p(-3 -1) q(-1,-7) and r(3,3) and points a and b are midpoints of segment pq and segment rq respectivel

Serhud [2]

Answer:

View graph

Step-by-step explanation:

Be: p(-3,-1) q(-1,-7) r(3,3)

we must find the value of the middle point a and b

$a=\frac{pq}{2}\\b=\frac{rq}{2}$

remember that distance between two points is:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$

pq=distance between p and q

qr =distance between q and r $pq=\sqrt{(-1+3)^{2}+(-7+1)^{2}}=\sqrt{(2)^{2}+(-6)^{2}}=\sqrt{40}\\ pq=2\sqrt{10}\\ a=\frac{pq}{2}=\frac{2\sqrt{10}}{2}=\sqrt{10}\\ coordinates\\a=\frac{pq}{2}=\frac{(-3-1,-7-1)}{2}=(-2,-4)\\\\qr=\sqrt{(3+1)^{2}+(3+7)^{2}}=\sqrt{(4)^{2}+(10)^{2}}=\sqrt{116}\\qr=2\sqrt{29}\\ b=\frac{qr}{2}=\frac{2\sqrt{29}}{2}=\sqrt{29}\\ coordinates\\b=\frac{pq}{2}=\frac{(-1+3,-7+3)}{2}=(1,-2)\\$