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2 years ago

What is the answer to 1014.16667 rounded to the nearest tenths

Mathematics

1 answer:

yan [13]2 years ago

6 0

Answer: 1014.2

Step-by-step explanation: you look to the right of 1 and if it is 5 or greater you round up. If it is 4 or less you keep it the same number as it is.

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Distributive Property Using GCF 25 + 30

postnew [5]

Hey there! I'm happy to help!

Let's look at the factors of each of these numbers.

1,25

5,5

1,30

2,15

3,10

5,6

The greatest common factor between these is 5. Therefore, we divide 25+30 by 5 and put the 5 on the outside.

5(5+6).

This has the same value but it is just written differently.

Have a wonderful day! :D

3 0

2 years ago

Solve 8[7-(6x-6)]+6x=0.

zavuch27 [327]

I believe that X would be -2.47619.
First you distribute the -1 in front of the parenthesis. Then distribute the 8 outside of the bracts. And finally just solve for X by combing like terms.

4 0

2 years ago

Which of the statements below best describes direct variation?

REY [17]

Direct variation is when one variable changes the other changes in proportion of the first, therefore all the above

8 0

3 years ago

The equation of a circle is given below.

BARSIC [14]

Given:

The equation of the circle is $x^2+(y+4)^2=64$

We need to determine the center and radius of the circle.

Center:

The general form of the equation of the circle is $(x-h)^2+(y-k)^2=r^2$

where (h,k) is the center of the circle and r is the radius.

Let us compare the general form of the equation of the circle with the given equation $x^2+(y+4)^2=64$ to determine the center.

The given equation can be written as,

$(x-0)^2+(y+4)^2=64$

Comparing the two equations, we get;

(h,k) = (0,-4)

Therefore, the center of the circle is (0,-4)

Radius:

Let us compare the general form of the equation of the circle with the given equation $x^2+(y+4)^2=64$ to determine the radius.

Hence, the given equation can be written as,

$x^2+(y+4)^2=8^2$

Comparing the two equation, we get;

$r^2=8^2$

$r=8$

Thus, the radius of the circle is 8

3 0

3 years ago

SOLVE BY ELIMINATION. x-2y=2 3x+y=6

USPshnik [31]

$\left \{ {{x-2y=2} \atop {3x+y=6\ |\cdot2}} \right. \\\\ \left \{ {{x-2y=2} \atop {6x+2y=12}} \right. \\+-----\\\\7x=14\ \ \ \ |:7\\x=2\\\\2-2y=2\\-2y=2-2\\-2y=0\\y=0\\\\ \left \{ {{y=0} \atop {x=2}} \right.$

7 0

2 years ago