Help meeee??!??!?! pleasee

A building contractor examined 6 floor joists in a home and found that they had the following widths:

miss Akunina [59]

Answer:

Step-by-step explanation:8 inches

89 is the answe 89

8 0

2 years ago

The aspect of geography that examines physical characteristics of the Earth is known as ______________.

antoniya [11.8K]

<span>The aspect of geography that examines physical characteristics of the Earth is known as physical geography. The correct option among all the options that are given in the question is the third option or option "C".

Number of people in the year 2010 = 308.7 million
Number of people in the year 2000 = 281.4 million
Number of people by which the population grew = (308.7 - 281.4) million
= 27.3 million
Percentage of change in the population = (27.3/308.7) * 100
= 8.84 percent.

</span>

5 0

3 years ago

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3. The values for p and q represent the equilibrium point. As related to the troupe's

svp [43]

Answer: The equilibrium point represents the raising or lowering the price in response to changes in the supply or demand.

If the price of a good is above equilibrium, this means that the quantity of the good supplied exceeds the quantity of the good demanded.

If the quantity is below the equilibrium point, it will create a shortage. because the quantity supplied is less than quantity demanded.

Hope this helps!

Step-by-step explanation:

5 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

I don’t know how to answer this question...

EleoNora [17]

Answer:

the pic is nit clear but nice pfp tanjiro from demon slayer

also if u don't mind me asking have you watched the mugen train

7 0

3 years ago

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