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3 years ago

Which of the following are solutions to the inequality below? select all that apply

1 answer:

6 0

The answer is #1, #2, and #4

It’s #1 because if you look at a number line -3 is before -4

It’s #2 because it’s less then or equal to and -4 is equal to -4

And lastly it’s #4 because 3 is greater then -4

Hope this helps!!

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Round 8.54951607694 to 6 decimal places.

Ratling [72]

Answer:

8·549516

Step-by-step explanation:

To round the number given to 6 decimal place, we will follow the steps below;

First count six digits after the decimal point

Then after the six digit number take the next number and see if it below 5 then this means you will be rounding down, so you will leave your six digit after the decimal points the way they are and discard the other digits after the the six digits, but if the digit number right after the sixth digit is 5 and above, then you will be rounding up, you will add one to your sixth digit and then discard the digits after the sixth digits.

That is;

In the number given: 8.54951607694

The sixth digit after the decimal point is 6, the number that comes after it is zero, so we will leave the 6 the way it is and discard the other digits after 6

8.54951607694 ≈ 8·549516 to 6 decimal place

4 0

3 years ago

2) When we divide ______ by 8 gives a quotient of 75 and a remainder of 3

Pani-rosa [81]

Answer:

603

Step-by-step explanation:

603/8

600/8 + 3/8 = 75, 3/8

4 0

3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

Spencer purchases clothing for the new school year. Polos cost $12.00 each, and jeans

pentagon [3]

Answer: Two pairs of jeans and three polo's

Step-by-step explanation:

N/A

3 0

3 years ago

A cylindrical tank has a base with a circumference of meters and an isosceles right triangle painted on the interior side of the

iren2701 [21]

C. Under root 2 there is an answer

7 0

3 years ago