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3 years ago

Write an equation to the model above BEST ANSWER GET BRANLIEST

Mathematics

1 answer:

erastovalidia [21]3 years ago

8 0

Answer:

0 ≤ x ≤ 2 (i assumed that the arrows included 0 and 2).

Step-by-step explanation:

One arrow shows 0 and greater (as it is pointing right) and the other shows 2 and smaller (as it is pointing left). The arrow pointing right has the equation of:

x ≥ 0 (where x is a random number)

The arrow pointing left has the equation of:

x ≤ 2 (where x is a random number)

Assuming that you need to find an equation so that x fits both inequalities, we combine them into one inequality and get:

0 ≤ x ≤ 2

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Tu has a midpoint at M(0,2) . point T is at (0,3) . Find the coordinates of point U.

Ne4ueva [31]

The answer will be/is 0,1

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3 years ago

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Answer:

-7/3x + 3

Step-by-step explanation:

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3 years ago

Read 2 more answers

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

Write an algebraic expression for each word expression. Then evaluate the expression for these values of the variable : 1/2,4,an

Yuki888 [10]

Answer:

1)p/4,2,1,1.625 2)x+1,3/2,5,7.5

Step-by-step explanation:

1) p/4

1/2 divided by 4=2

4/4=1

6.5/4=1.625

2) x+1

1/2+1=3/2

4+1=5

6.5+1=7.5

7 0

3 years ago

On May 14, 1898, a severe hailstorm hit Kansas City. The largest hailstones were 9.5 inches in circumference. Find the radius of

tigry1 [53]

C = $\pi$ r
9.5 = 2 $\pi$ r
$\pi$ r = 4.75
r = 1.511971959
The number above should be rounded to the nearest 2 decimal places, so the answer is 1.51.

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3 years ago