All categories

2 years ago

Pablo made $357 for 17 hours of work.

Mathematics

1 answer:

vredina [299]2 years ago

3 0

Answer:

Step-by-step explanation:

357÷17=21

168÷21=8

so $357 divided by the 17 hours is 21

then you do the $21 per hour multiplied by the money is what he makes then divide 21 and 168

You might be interested in

How to do this problem because it's hard

Elenna [48]

You will need to set up and solve 2 equations:
A) 3 Vans + 10 Buses = 379
B) 14Vans + 13 Buses = 624
Multiply equation A by -14
A) -42 Vans -140 Buses = -5,306 then Multiply equation B by 3
B) 42 Vans + 39 Buses = 1,872
Adding BOTH equations
-101 Buses = -3,434
Each Bus holds 34 Students

Putting this into equation A
A) 3 Vans + 10 Buses = 379
A) 3 Vans + 10 * 34 = 379
A) 3 Vans +340 = 379
A) 3 Vans = 39
Each van = 13 students

4 0

3 years ago

ILL BRAINLIEST YOU PLEASE HELP ME

astra-53 [7]

Answer:

a) 5.6

Step-by-step explanation:

sin 22° = x/15

0.3746 = x/15

x = 5.62

3 0

3 years ago

How do how you cross cancel multiplying fractions

DENIUS [597]

It's easy once you spot the ones that can cross cancel!

Say we have the fractions 8/10 and 20/23. $\frac{8}{10} \frac{20}{23}$ (it's easier to see on top of each other)

If you look diagonally , so 8 and 23 and 10 and 20, you can see that 10 and 20 have a common factor. So we divide it by the highest common factor to reduce those numbers, making it easier to multiply. 10 and 20 can become 1 and 2, dividing by 10. So now we are left with 8/1 and 2/23, and now we multiply normally going across so 16/23.

This works going both diagonals and simplifying both, but in that case it would be easier to try and simplify the fractions before cross multiplying them.

Basically: look for those diagonals and if they can be divided down by the highest common factor, go for it to make it easier to multiply normally afterwards.

Hope I helped!

5 0

3 years ago

Chamblee High School is selling Valentine's Day gifts as a fundraising event. One long stemmed rose costs $3.00 while one long s

Mumz [18]

Answer: Choice B) 60 roses and 10 carnations

============================================================

Explanation:

r = number of roses
c = number of carnations

r and c are positive whole numbers.

r+c = total number of flowers = 50, since 50 orders are made.

The first equation to set up is r+c = 50.

This equation can be solved to get r = 50-c.

------------------

3r = cost of all the roses only, in dollars

1.5c = cost of all the carnations only, in dollars

3r+1.5c = total cost of all the flowers = 195 dollars

3r+1.5c = 195

------------------

Let's apply substitution to solve

3r+1.5c = 195

3(50-c)+1.5c = 195

150-3c+1.5c = 195

-1.5c+150 = 195

-1.5c = 195-150

-1.5c = 45

c = 45/(-1.5)

c = -30

That's not good. We shouldn't get a negative value.

It turns out that the condition r+c = 50 should be ignored. Notice how none of the answer choices listed have r+c leading to 50.

So we'll only focus on the equation 3r+1.5c = 195

-----------------

If we plugged in r = 100 and c = 100, then we get

3r+1.5c = 195

3(100)+1.5(100) = 195

300+150 = 195

450 = 195

Which is false. So we can rule out choice A

Let's repeat those steps for choice B

3r+1.5c = 195

3(60)+1.5(10) = 195

180 + 15 = 195

195 = 195

So that works out. I have a feeling your teacher meant to say "70 orders" instead of "50 orders". If so, then the equation r+c = 50 would be r+c = 70 and everything would lead to choice B as the final answer.

Choices C and D are similar to that of choice A, so they can be ruled out.

8 0

3 years ago

Graph the line given by 2x+y=1 and the circle given by x²+y²=10.Find all solutions to the system of equations. Verify your resul

Aleksandr-060686 [28]

Answer:

(1.8, -2.6) and (-1, 3)

Step-by-step explanation:

$2x+y=1$

$x^2+y^2=10$

From the first equation

$y=1-2x$

Applying to the second equation

$x^2+(1-2x)^2=10\\\Rightarrow x^2+1+4x^2-4x=10\\\Rightarrow 5x^2-4x-8=0$

Solving the equation we get

$x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\cdot \:5\left(-9\right)}}{2\cdot \:5}, \frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:5\left(-9\right)}}{2\cdot \:5}\\\Rightarrow x=1.8, -1$

At x = 1.8

Applying in first equation

$2\times 1.8+y=1\\\Rightarrow y=1-3.6\\\Rightarrow y=-2.6$

At x = -1

Applying in first equation

$2\times -1+y=1\\\Rightarrow y=1+2\\\Rightarrow y=3$

∴ The circle and line intersect at points (1.8, -2.6) and (-1, 3)

3 0

3 years ago