All categories

2 years ago

Can somebody check to see if these are right? THANK YOU!!

Mathematics

2 answers:

PIT_PIT [208]2 years ago

8 0

The first one should be x - y <= 25. Since there are 25 OR fewer available, depending on who is out sick or on vacation.

Fed [463]2 years ago

3 0

Those answers are correct cause i did that before.

Hope this Helped!

;D
Brainliest??

You might be interested in

Find the solution of the system of equations.

AnnyKZ [126]

Answer:

x=-2 and y=-3

Step-by-step explanation:

2x+4y=-16

-2(-2x+2y=-2)

2x+4y=-16

4x-4y=4

6x=-12

x=-2

2x+4y=-16

2(-2)+4y=-16

-4+4y=-16

+4 +4

4y=-12

y=-3

6 0

2 years ago

Read 2 more answers

Find the diameter of a circle with an area of 6767 square centimeters.

scZoUnD [109]

Area=6767cm²

Diameter be d

Radius be r

πr²=6767
r²=6767/π
r²=2155
r=46.4cm

Diameter=2r=46.4(2)=92.8cm

6 0

1 year ago

Read 2 more answers

*pls help* <br> solve 7,8,9, and 10 for brainliest

ivanzaharov [21]

Answer:

#8 : x=15−y

#9: $\frac{c}{b} -\frac{ax}{b}$

#10: y=2A−x

#7: T= $\frac{P}{IR}$

Step-by-step explanation:

7 0

2 years ago

Laura determined the difference between 8 and –2 using the additive inverse. Her steps are shown below.

NNADVOKAT [17]

Answer: The 8 should have remained positive.

6 0

2 years ago

Read 2 more answers

The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

7 0

3 years ago