Consider the system of equations.

1 answer:

4 0

We know that
The subtraction property of equality<span> tells us that if we subtract from one side of an equation, we also must subtract from the other side of the equation to keep the equation the same

that means

if a=b and c=d
a-c=b-d

</span>then
<span>−8x+4y=0----------> equation 1
a=(-8x+4y) b=0

−8x+7y=6----------> equation 2
c=(-8x+7y) d=6

therefore
</span>
(-8x+4y)-(-8x+7y)=0-6
-8x+4y+8x-7y=-6
-3y=-6

the answer is the option
Subtraction Property of Equality

You might be interested in

What are the coordinates of the point on the directed line segment from K (-5, -4) to L (5,1) that partitions the segment into a

Dima020 [189]

Answer:

i think its d

Step-by-step explanation:

3 0

3 years ago

Can someone help me please

ss7ja [257]

Answer:I think it’s c

Step-by-step explanation:

3 0

3 years ago

Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$