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3 years ago

Simplify the expression8(5 plus w)

Mathematics

1 answer:

inna [77]3 years ago

8 0

8(5 plus w)

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8*5 + 8*w

40 + 8w

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Oceanside Bike Rental Shop charges a 13 dollar fixed fee plus 9 dollars an

tatyana61 [14]

58-13=45 45/9=5 Carlos rented the bike for 5 hours

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4 years ago

Read 2 more answers

Determine which of the following terms are not considered to be like terms with the expression -6s ( the two is squared like on

Iteru [2.4K]

Answer:

$s^2t^2$

$4(s\cdot t)$

$-6(s^2+t)$

Step-by-step explanation:

We want to select all the terms that are not considered to be like terms with $-6s^2t$ .

The terms that are like terms with $-6s^2t$ must have $s^2t$ .

It doesn't matter the coefficient.

So we can easily see that all the following are not like terms with $-6s^2t$ :

$s^2t^2$

$4(s\cdot t)$

$-6(s^2+t)$

4 0

3 years ago

25 points America explain

Gnesinka [82]

Answer:

Degree of length: 1

Degree of width: 1

Degree of height: 3

Degree of volume: 5

Explanation:

The degree, for example, is the 2 in 3x²-1 or the 3 in x³·5x²-4. Because the highest exponent in the equation determines the degree or something like that so yea i got u fam.

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3 years ago

Find the slip of every line that is parallel to the graph of the equation x-8y=-1

LenaWriter [7]

Answer:

1/8

Step-by-step explanation:

x-8y=-1

First, put the equation in slope intercept form ( y= mx+b where m is the slope and b is the y intercept)

-8y = -x-1

y = 1/8x +1/8

The slope is 1/8 and the y intercept is 1/8

The slope for parallel lines is the same for all parallel lines

1/8

8 0

3 years ago

Inverse notation f^-1 used In a pure mathematics problem is not always used when finding inverses of applied problems. Rather, t

solong [7]

Answer:

a) 75.4 kg

b) $h(W)=\frac{W+83}{2.2}$

Step-by-step explanation:

a) The ideal weight of a 6-foot (72 inches) male is given by simply applying h= 72 in to the expression:

$W(72) = 49+2.2(72-60)\\W(72) =75.4\ kg$

b) Expressing height as a function of weight:

$W(h)= 49+2.2(h-60)\\2.2h-132+49=W\\h(W)=\frac{W+83}{2.2}$

Verifying with W(h(W)):

$W(h(W))= 49+2.2(\frac{W+83}{2.2} -60)\\W(h(W))= 49-132+W+83\\W(h(W))=W$

Verifying with h(W(h):

$h(W(h))=\frac{(49+2.2(h-60))+83}{2.2}\\h(W(h))=\frac{(49+2.2h-132+83)}{2.2}\\h(W(h))=h$

8 0

3 years ago