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3 years ago

I need help on this math problem !

Mathematics

1 answer:

xeze [42]3 years ago

8 0

Answer:

181 units²

Step-by-step explanation:

Let's break up the polygon into 3 shapes

Step 1: Area of the left rectangle

4(7) = 28 units²

Step 2: Area of middle rectangle

9(13) = 117 units²

Step 3: Area of top square

6² = 36 units²

Step 4: Add them all up

28 + 117 + 36 = 181 units²

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[2/SEC(¶/3)•[lim x→0 x^3+8x+10]^2]/[lim θ→0 sinθ/θ]

Oliga [24]

Assuming the pilcrow is supposed to represent pi:

$\frac{\frac{2}{sec(\frac{\pi}{3})}*(\lim_{x \to 0} x^3+8x+10)^2}{ \lim_{\theta \to 0} \frac{sine(\theta)}{\theta}}\\\frac{\frac{2}{\frac{1}{cosine(\frac{\pi}{3})}}*((0)^3+8(0)+10)^2}{\frac{sine((0))}{(0)}}\\\frac{\frac{2}{\frac{1}{\frac{1}{2}}}*(0+0+10)^2}{\frac{0}{0}}\\\frac{\frac{2}{2}*(10)^2}{ \lim_{\theta \to 0} \frac{cosine(\theta)}{1}}\\\frac{100}{cosine((0))}\\\frac{100}{1}\\100$

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3 years ago

There are 3 marbles (yellow,blue and green). You pick a marble and flip a coin.

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1/3, bc you have 33.333% for each marble that it would be chosen.

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3 years ago

Read 2 more answers

If Randy sells 8 times as many vacuum cleaners as Janice, and Janice sells 690 vacuum cleaners per year, on average, how many do

drek231 [11]

Randy sells, in average, 460 vacuum cleaners per month.

<h3>How many vacuum cleaners does Randy sell each month?.</h3>

First, we know that Janice sells 690 units per year.

There are 12 months in a year, so the amount that she sells, in average, per month is:

J = 690/12 = 57.5

This means that Janice sells, in average, 57.5 units per month. And Randy sells 8 times as many, then we can solve the product:

R = 8*57.5 = 460

This means that Randy sells 460 units per month.

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2 years ago

How to solve -13=5(1 4m)-2m?

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First, you would multiply: -13 = 70m - 2m. Then, you subtract the like terms, which would be 70m - 2m: -13 = 68m. You are looking for what m is, so you would divide 68 on both sides: m = - 13 / 68. In math, people would usually like fractions more than decimals, but if you are looking for the decimal number, it would be m ≈ - 0.19.

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3 years ago

The first two terms in an arithmetic progression are -2 and 5. The last term in the progression is the only number in the progre

Rudik [331]

Given:

The first two terms in an arithmetic progression are -2 and 5.

The last term in the progression is the only number in the progression that is greater than 200.

To find:

The sum of all the terms in the progression.

Solution:

We have,

First term : $a=-2$