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4 years ago

Can someone please answer. There is a picture. It is only one question. Thank you so much!!!

Mathematics

1 answer:

dalvyx [7]4 years ago

6 0

Is the second one
x^2+(y-2)^2=25

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Solve the following equation for y. Y= 2y+4y-3=27

sammy [17]

Answer:

$2y + 4y - 3 = 27 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 6y = 27 + 3 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 6y = 30 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = \frac{30}{6} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = 5$

4 0

3 years ago

This is due today !! help pls

Anarel [89]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Suppose you received a score of 95 out of 100 on exam 1 . The mean was 79 and the standard deviation was 8 . If your score on ex

Schach [20]

Answer:

You did the same on both exams.

Step-by-step explanation:

To compare both the scores, we need to compute the z scores of both the exams and then compare the values. The formula for z-score is:

Z = (X - μ)/σ

Where X = score obtained

μ = mean score

σ = standard deviation

For Exam 1:

Z = (95 - 79)/8

= 16/8

Z = 2

For Exam 2:

Z = (90 - 60)/15

= 30/15

Z = 2

The z-scores for both the tests are same hence the third option is correct i.e. you did the same on both exams.

8 0

3 years ago

How to factor 25x^2 - 81= 0

dalvyx [7]

Your question is a difference of squares. both terms that need to be factored are squared. to solve take the square root of each put into two terms one of addition and one of subtraction.

anwser= (5x-9)(5x+9)

the reason this works is because when you foil you get
25x²-45x+45x-81
the middle terms cancel revealing
25x²-81

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