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mash [69]
3 years ago
15

1(whole number) 4/6 (fraction) + 2(Whole number) 1/4 (fraction)

Mathematics
2 answers:
ivanzaharov [21]3 years ago
5 0

Answer:

Step-by-step explanation:

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lbvjy [14]3 years ago
4 0

Answer:

answer is 3 11/12

Step-by-step explanation:

1 4/6 =10/6

2 1/4=9/4

when adding fractions that are do not share a common denominatour you have to find a common multiple

which in this case is 12

so 9/4 timesed the top and the bottom by 3 gives you 27/12

and 10/6 timesed the top and bottom by 2 gives you 20/12 and then you add the the nominators which is 20+27 which then gives you 47/12 and then to change it back to mixed fraction you will get 3 (wholes) and 11/12 (fraction)

Hope this helps xx

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Is 12 a solution to the equation × - 8 = 20? Explain how you know.
jeyben [28]

We have the following:

x-8=20

solving for x

\begin{gathered} x=20+8 \\ x=28 \end{gathered}

The solution of x is equal to 28, therefore 12 is not a solution of the equation

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1 year ago
For f(x) = 4x +1 and g(x) = x2 -5, find (f 3)(x).
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D!! It’s super easy I hope you can learn how to do it
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What is the balance on a deposit of $1,150 that earns 9% interest compounded annually for 2 years
igor_vitrenko [27]
Step One: Calculate interest.

1150*0.09=103.5

Step Two: Calculate the amount at the end of the first year.

1150+103.5=1253.5

Step Three: Calculate the amount at the end of the second year. This will be your final answer.

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3 0
3 years ago
Square root of 120 over the square root of 30. what is the following quotient.
Minchanka [31]

Answer:

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Step-by-step explanation:

3 0
1 year 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
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