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3 years ago

Please help me with this 10 points

Mathematics

1 answer:

kiruha [24]3 years ago

4 0

Answer:

i couldnt see the first 2 good.

3. exact form= 58/15 decimal form= 3.86 mixed # form= 3 13/15

4. exact form= 13/8 decimal form = 1.625 mixed numbers 1 5/8

5.convert thw mixed number to improper fractions, then find the LCD and combine...-8

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Is 0 an integer ?<br> It wont let me post without making it longer

lina2011 [118]

Yes, it is. You may prove it using something like this: 1 is an integer. 1-1 is a difference between integers. A difference between integers returns an integer, so 0 is an integer.

7 0

3 years ago

Determine the measure of each segment then indicate whether the statements are true or false

kupik [55]

Answer:

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

Step-by-step explanation:

Considering the graph

Given the vertices of the segment AB

A(-4, 4)

B(2, 5)

Finding the length of AB using the formula

$d_{AB}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(2-\left(-4\right)\right)^2+\left(5-4\right)^2}$

$=\sqrt{\left(2+4\right)^2+\left(5-4\right)^2}$

$=\sqrt{6^2+1}$

$=\sqrt{36+1}$

$=\sqrt{37}$

$d_{AB}\:=\sqrt{37}$

$d_{AB}=6.08$ units

Given the vertices of the segment JK

J(2, 2)

K(7, 2)

From the graph, it is clear that the length of JK = 5 units

$d_{JK}=5$ units

Given the vertices of the segment GH

G(-5, -2)

H(-2, -2)

Finding the length of GH using the formula

$d_{GH}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(-2-\left(-5\right)\right)^2+\left(-2-\left(-2\right)\right)^2}$

$=\sqrt{\left(5-2\right)^2+\left(2-2\right)^2}$

$=\sqrt{3^2+0}$

$=\sqrt{3^2}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

$d_{GH}\:=\:3$ units

Thus, from the calculations, it is clear that:

$d_{AB}=6.08$

$d_{JK}=5$

$d_{GH}\:=\:3$

Thus,

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

8 0

3 years ago

Can somebody help. 7 out of 10 is what percent?

Otrada [13]

Answer:

70%

Step-by-step explanation:

Multiply by ten and you get your percent

3 0

3 years ago

Read 2 more answers

In great detail, with full work done out, please find x<br> 1 + x = 2

kenny6666 [7]

Answer:

x = 1

Step-by-step explanation:

1 + x = 2

Subtract 1 form both sides,

1 + x - 1 = 2 - 1

x = 1

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

Please help me with this 10 points​

Please help me with this 10 points