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3 years ago

Helpp mee pleaseeeeeeee

Mathematics

1 answer:

mario62 [17]3 years ago

8 0

Answer:

Step-by-step explanation:

2((8)+(12)+(6))

2(8+12+6)

2(26)

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Why did Dante write the Inferno

makkiz [27]

Answer:

he used it as a vehicle to express his political beliefs and take comfort in imagining bad ends for his enemies. However, the poem's main purpose is, to quote Milton, to "justify the ways of G0d to Men."

Step-by-step explanation:

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3 years ago

Which expression is equivalent to 3 (4x - 5)?

Rufina [12.5K]

The answer is D 12x-15

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3 years ago

To the nearest tenth of a degree, ﬁnd the sizes of the acute angles in the 7-24-25 right

fiasKO [112]

First of all, just to avoid being snookered by a trick question, we should verify that these are really right triangles:

7² + 24² really is 25² , and 8² + 15² really is 17² , so we're OK there.

In the first one:
sin(one acute angle) = 7/25 = 0.28
the angle = sin⁻¹ (0.28) = 16.26°
the other acute angle = (90° - 16.26°) = 73.74°

In the second one:
sin(one acute angle) = 8/17 = 0.4706...
the angle = sin⁻¹ (0.4706...) = 28.07°
the other acute angle = (90° - 28.07°) = 61.93°

I'm sorry, but just now, I don't know how to do the
third triangle in the question.

7 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

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3 years ago

The lines are parallel. Find the slope-intercept form of the equation y₂.

Bas_tet [7]

Since the lines are parallel and you know the slope of $y_{1}$ is -2, you know the slope of line $y_{2}$ will also be -2. You can see on the coordinate grid that line $y_{2}$ crosses the y-axis at -1. When you piece all of these facts together into slope-intercept form you get $y_{2}$ = -2x - 1.

6 0

3 years ago