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2 years ago

I need help please anyone??

Mathematics

2 answers:

lys-0071 [83]2 years ago

5 0

Answer:

A. 26.23

Step-by-step explanation:

wel2 years ago

3 0

Answer:

A.$26.23

Step-by-step explanation:

First, you subtract $311.12 from$ 835.72

$835.72+$311.12=$524.6

Then, you split the $524.6 by devonte and the 19 other kids.

$524.6/20=$26.225

Finnally you round the quotient to the ten-hundrenths place.

Your final answer should be $26.23.

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alekssr [168]

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C. 20

Step-by-step explanation:

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2 years ago

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Igoryamba

The equation does not fit the table

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3 years ago

#3) Simplify: 5(j - 3) - (2j + 2) I might ask for help in a bit just plz help I’m not good with math

Vesna [10]

Answer:

5j-15-2j+2= 3j-13

Step-by-step explanation:

5 times j

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2 years ago

Complete the sentence by selecting the correct phrase. The expression (a2−)2−(b2)2 is an example of: a system of equations, a po

ss7ja [257]

Answer:

a difference of squares

Step-by-step explanation:

Let's write down the expression:

$(a^2)^2 -(b^2)^2$

we can see that this expression is one of the form $x^2 -y^2$ which is the general expression of a difference of squares.

To be more specific and to see why we say this, let's make $x=a^2$ and $y=b^2$

Now, if we substitute this in the general formula, we get:

$x^2 -y^2 =(a^2)^2 -(b^2)^2$

Thus, this expression is a difference of squares.

4 0

3 years ago

A rubber ball dropped on a hard surface takes a sequence of bounces, each one 1/6 as high as the preceding one. If this ball is

Brums [2.3K]

Answer:

$\frac{259}{54}\text{ or }4.8\text{feet}$

Step-by-step explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one $\frac{1}{6}$ as high as the preceding one.

TO FIND: If this ball is dropped from a height of $12$ feet, how far will it have traveled when it hits the surface the fifth time.

SOLUTION:

once the ball is dropped on hard surface it bounces $\frac{1}{6}$ of preceding one and comes down the same distance.

When the ball is dropped from $12$ feet height

after first hit $=12\times\frac{1}{6}\text{ up}+12\times\frac{1}{6}\text{ down}=4$

new height $=12\times\frac{1}{6}=2\text{feet}$

after second hit $=2\times\frac{1}{6}\text{ up}+2\times\frac{1}{6}\text{ down}=\frac{2}{3}$

new height $=2\times\frac{1}{6}=\frac{1}{3}\text{ feet}$

after third hit $=\frac{1}{3}\times\frac{1}{6}\text{ up}+\frac{1}{3}\times\frac{1}{6}\text{ down}=\frac{1}{9}$

new height $=\frac{1}{3}\times\frac{1}{6}=\frac{1}{18}\text{ feet}$

after fourth hit $=\frac{1}{18}\times\frac{1}{6}\text{ up}+\frac{1}{18}\times\frac{1}{6}\text{ down}=\frac{1}{54}$

adding all distance $=4+\frac{2}{3}+\frac{1}{9}+\frac{1}{54}$

$=\frac{259}{54}$ feet

Hence the ball will travel $\frac{259}{54}$ feet before it hits the surface fifth time.

4 0

3 years ago