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babymother [125]
3 years ago
5

Write the expression as a single power using only positive exponents, if possible. Assume no denominator equals zero. Will give

brainliest!

Mathematics
1 answer:
Olenka [21]3 years ago
4 0
The answer is 1/x^10.

You subtract exponents when you divide, so: x^-6 - x^4. That equals x^-10.

Since the exponent is negative, you must put it in the denominator. Since this fraction has nothing in the numerator, use "1" as a placeholder.

Therefore, the answer is 1/x^10. Hope this helps!
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Rama09 [41]

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about an hour and 42 minutes

Step-by-step explanation:

Time=Distance/Speed

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Kenya dose the pole vault for her school's track team. Her record is 2 meters 86 centimetres. How many millimetres is Kenya's re
kvv77 [185]

Answer:

2860

Step-by-step explanation:

First, convert 2 meters into cm and then convert to mm. m to cm+200cm cm-mm= 2000. Then turn 86 centimetres into mm which is 860 mm. Lastly 2000+860=2860. Hope this helps

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The length of a rectangle is 3 meters more than twice its width.The perimeter of the rectangle is 48 meters. Let W represent the
lord [1]
Perimeter of a rectangle: 2(l+w)

Let width be 'x'

Length is 3 meters more than twice 'x'
Length= 2x+3.

L= x
W = 2x+3


Formula =
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8 0
4 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
The difference between 3 times a number x and 2 is 19. What is the value of x? ​
Lisa [10]

Answer:

The answer is 7

7 0
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