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3 years ago

Help me it’s angle measurement

Mathematics

1 answer:

Dmitriy789 [7]3 years ago

4 0

The first question is c and the second one is a
Ik the first one is c but the second one I think is that

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Three times two less than a number is greater than or equal to five times the number. Find all of the numbers that satisfy the g

Sholpan [36]

It would be 3(n-2) is greater than or equal to 5n.

3 0

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

4 years ago

Re-write the quadratic function below in Standard Form y = -(x – 3)2 +8

lisov135 [29]

Answer:

y = -x² + 6x - 1

General Formulas and Concepts:

Algebra I

Standard Form: ax² + bx + c = 0
Expanding by FOIL (First Inside Outside Last)
Combining like terms

Step-by-step explanation:

Step 1: Define equation

y = -(x - 3)² + 8

Step 2: Rewrite

Expand [FOIL]: y = -(x² - 6x + 9) + 8
Distribute -1: y = -x² + 6x - 9 + 8
Combine like terms: y = -x² + 6x - 1

4 0

3 years ago

Explain what is the 7th term in the geometric sequence described by this explicit formula? A.-2187 B.6561 C.-6561 D.2187

Shtirlitz [24]

Answer:

the answer is d

Step-by-step explanation:

5 0

3 years ago

Find each product.

kvasek [131]

Answer: The answer is -2

Step-by-step explanation: positive and a positive is a positive, but, a positive and a negative is a negative.

3 0

4 years ago