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2 years ago

Help meee please 20 points

Mathematics

2 answers:

Agata [3.3K]2 years ago

7 0

The answer is sixteen

Lyrx [107]2 years ago

3 0

Answer:

16 sq. yds

Step-by-step explanation:

1. Split the shape into 2

The small square is 2(2) = 4

The bigger rectangle is 3(4) = 12

4 + 12 = 16 sq. yds

You might be interested in

The equation of the graphed line is 2x – 3y = 12.

bearhunter [10]

Hello there! The x-intercept is 6.

The x-intercept is when the y value id equal to zero, and the line crosses the x axis. In this example, you can see that the line passes through x at the value of 6, making that your answer.

I hope this was helpful and have a great rest of your day!

4 0

3 years ago

Read 2 more answers

A coffee business sells a pound of coffee for $ 9.75. Monthly expenses are $ 4,500 plus $ 4.25 for every pound of coffee sold. a

hram777 [196]

Answer:

a. IT = 9.75 * X

b. GT = 4500 + 4.25 * X

c. G = 9.75 * X - 4500 - 4.25 * X

Step-by-step explanation:

With the data of the statement we can get a function. Let X be the number of pounds sold.

to. Monthly income

IT = 9.75 * X

b. Monthly expenses

GT = 4500 + 4.25 * X

c. Monthly Earnings (Monthly Income - Monthly Expenses)

G = 9.75 * X - (4500 + 4.25 * X)

G = 9.75 * X - 4500 - 4.25 * X

4 0

3 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

3 years ago

Which of the following points lies in the solution set of the inequality y > 3x +10?

shepuryov [24]

Since 3 is greater than -3, hence (-1, 3) lie in the solution set. Option C is correct

In order to determine the points that lie in the solution set of the inequality y > 3x +10, we will substitute the x-coordinate and see if <u>y is greater than the result.</u>

<u />

For the coordinate point (1, 10)

y > 3(1) +10

y > 13

Since 10 is not greater than 13, hence (1,10) does not lie in the solution set.

For the coordinate point (4, 20)

y > 3(4) +10

y > 22

Since 20 is not greater than 22, hence (4,20) does not lie in the solution set.

For the coordinate point (-1, 3)

y > 3(-1) +10

y > -7

Since 3 is greater than -3, hence (-1, 3) lie in the solution set.

Learn more on inequality here: brainly.com/question/24372553

7 0

3 years ago

A gym membership charges $50 new member fee and $24 for every month after. Write an equation for this where y is the total cost

DanielleElmas [232]

It's 202739267166262

6 0

3 years ago

Read 2 more answers