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4 years ago

If 12 and 20 each divide K without a remainder, what is the value of K?

Mathematics

2 answers:

Sidana [21]4 years ago

6 0

Answer:

Step-by-step explanation:

1*12=12 1*20=20

2*6=12 2*10=20

3*4=12 4*5=20

both have a similar multiple of 4, so k must be 4.

Juli2301 [7.4K]4 years ago

4 0

Factors of 12 : 1, 2, 3, 4, 6, 12

Factors of 20: 1, 2, 4, 5, 10, 20

In both lists of factors, 4 is in common. Therefore, the value of K is 4 because it can divide into both 12 and 20 without a remainder. 2 is also another possible value of K.

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In the diagram, which pair of angles are vertical angles?

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Angles 1 and 3, angles 2 and 4, angles 5 and 7, and angles 6 and 8. They are also congruent, because vertical angles are congruent.

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3 years ago

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almond37 [142]

Answer:

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Step-by-step explanation:

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2 years ago

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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

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3 years ago

A rental car company charges $30 a day plus $0.05 a mile. Which expression could be used to find the cost of renting a car for m

Gnoma [55]

Answer:

Step-by-step explanation:

Since the company earns $0.05 for every mile, we would multiply m by 0.05. We also add it with 30 since it is a base charge. This makes it C

It can't be A or B since that would mean the company earns $30.05 for every mile.

It can't be D since that would mean the company earns $30 for every mile with a base charge of $0.05.

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3 years ago

Find the mode of the data set: 37, 42, 39, 44, 47, 38, 42, 45, 49, 35

Snezhnost [94]

Answer:

Step-by-step explanation:

mode is the most common number

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4 years ago

Read 2 more answers