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Marina CMI [18]

3 years ago

8

Joe got a bonus of $1,600.20 based on a certain percent of his sales for the year, which was $22,860. What percent of his sales

for the year was the bonus?

1 answer:

devlian [24]3 years ago

3 0

0.07 hope this helps

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a woman spent two thirds of her money. she lost two thirds of the remiander and then had $4(aus money) left. how much money did

antiseptic1488 [7]

Step-by-step explanation:

2/3 lost is 1/3 kept

2/3 of the 1/3 kept is 2/9 lost

1/3 x - 2/9 x = 4

3/9 x - 2/9 x = 4

1/9 x = 4

x = 36

First she lost 2/3 of 36, which is 24. She had 12 left. Then she lost 2/3 of 12 which is 8. She had 4 left.

8 0

2 years ago

Jose takes 120 minutes to weed 6 equal rows of vegetable plants in his garden. What is his unit rate

Zarrin [17]

Answer:

3 rows/hour

Step-by-step explanation:

It is given that, Jose takes 120 minutes to weed 6 equal rows of vegetable plants in his garden.

120 minutes = 2 hours

We need to find the unit rate per hour for weeding these rows of his garden. It can be calculated as follows :

$R=\dfrac{6}{2}\\\\R=3\ \text{rows/hour}$

So, the rate for weeding these rows of his garden is 3 rows per hour.

6 0

3 years ago

F(x) = 6x + 7; x = -3

jok3333 [9.3K]

Answer:

1+80×70-80+55-6+12065

8 0

3 years ago

Read 2 more answers

What is the following product

dangina [55]

Option B is correct.

Step-by-step explanation:

We need to solve: $\sqrt[3]{x^2}\sqrt[4]{x^3}$

We know that: $\sqrt[n]{x}\sqrt[b]{x} =\sqrt[n*b]{x.x}= \sqrt[n*b]{x^2}$

Applying the above rule:

$\sqrt[3]{x^2}\sqrt[4]{x^3}\\=\sqrt[3*4]{x^2.x^3}\\=\sqrt[12]{x^5}$

So, Option B is correct.

Keywords: Solving with Exponents

Learn more about Solving with Exponents at:

#learnwithBrainly

3 0

3 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

3 years ago

Read 2 more answers