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1 year ago

(5 - 7i)(8 + i)

Mathematics

1 answer:

docker41 [41]1 year ago

7 0

Answer:

(5 - 7i)(8 + i)

F) 40 - 561

H) 37 - 371

K) -16 - 2i

G) -33 - 61i

J) 47 - 51i

- 9 below

Step-by-step explanation:

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What other information do you need in order to prove the triangles congruent using the SAS congruence postulate?

ziro4ka [17]

Answer:

I think D

Step-by-step explanation:

I think that the answer is D, since the angles are not visibly shown to be equal. If you knew answer D, you would then be able to see that segments BC and CD are congruent. I am not 100% sure, however.

4 0

2 years ago

What’s the area of this figure

kvasek [131]

Answer:

743.25m^2

Step-by-step explanation:

Area of triangle = 1/2bh

= 1/2 x 30 x 26

= 390

Area of semi circle = 1/2 πr^2

= 1/ 2 x 3.14 x 15 x15

=353.25m^2

= 390 + 353.25

= 743.25m^2

8 0

2 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

2 years ago

To the nearest hundredth, what is the value of x?

Anna [14]

Answer:

1. We already have the measure of the hypotenuse and one out of two acute angle, therefore:

sin53° = x/45 => x = sin53° · 45 ≈35.94

2. We already have one out of two legs of the triangle and one acute angle so we know that:

tan27° = 48/x => x = 48/tan27° ≈ 94.21

3 0

2 years ago

Help (20 points, due tmr!! )<br><br> *pic below* :)

sveta [45]

Angles:
2 acute angles and 2 obtuse angles

Sides:
2 pairs pf parallel sides

7 0

2 years ago