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2 years ago

6

I need help thank youuu

1 answer:

WITCHER [35]2 years ago

6 0

Answer:

.

Step-by-step explanation:

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Which problem could be solved with the expression 4*(4+2)

Rina8888 [55]

The answer is A.

He ran for six (6) miles every day. 4 in the morning and 2 at night = 6 per day.

He did this for four (4) days. The way to figure this would be to go 6 * 4. This equals 24 miles.

Go back a few steps (punny!).

So days multiplied by how many miles he ran a day which is calculated by miles in morning added to miles at night.

days*(miles per morning + miles per night)

days=4
miles per morning=4
miles per night = 2

4 * (4+2)

Calculating this out....

4 * (4+2)

(4+2 = 6)

4 * 6 = 24.

So A is the answer.

3 0

3 years ago

Find the nth term of this sequence : 7 , 12 , 17 , 22 , 27 ???

spayn [35]

32 is the answer and it is adding 5 everytime

4 0

3 years ago

Read 2 more answers

Which expression is not equivalent to the other three?

AVprozaik [17]

Answer:

<em>4 + 7x – 2</em>

Step-by-step explanation:

<u>Equivalent Expressions</u>

Let's find out which expressions are equivalent by simplifying them. One of the expressions should be not equivalent to the other three.

Simplifying

$2x + 2+ 2x + 3x -8=7x-6$

Now for the second expression

$4 + 7x -2=7x+2$

The third expression simplifies to

$-2 + 5x + 2x - 4=7x-6$

Finally, this is the simplification for the expression

$8x - x - 6=7x-6$

As seen, the only expression not equivalent to the other three is 4 + 7x – 2

8 0

3 years ago

Read 2 more answers

Brian buys candy that costs 4 per pound. He will buy at least 8 pounds of candy. What are the possible amounts he will spend on

s2008m [1.1K]

Answer:

32 pound but I dunno what that is in american dollars

8 0

2 years ago

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Geometric Series assistance

Levart [38]

we have been asked to find the sum of the series

$\sum _{n=1}^5\left(\frac{1}{3}\right)^{n-1}$

As we know that a geometric series has a constant ratio "r" and it is defined as

$r=\frac{a_{n+1}}{a_n}=\frac{\left(\frac{1}{3}\right)^{\left(n+1\right)-1}}{\left(\frac{1}{3}\right)^{n-1}}=\frac{1}{3}$

The first term of the series is $a_1=\left(\frac{1}{3}\right)^{1-1}=1$

Geometric series sum formula is

$S_n=a_1\frac{1-r^n}{1-r}$

Plugin the values we get

$S_5=1\cdot \frac{1-\left(\frac{1}{3}\right)^5}{1-\frac{1}{3}}$

On simplification we get

$S_5=\frac{121}{81}$

Hence the sum of the given series is $\frac{121}{81}$

5 0

2 years ago