All categories

3 years ago

If the result of multiplying a number by 8/3 and then dividing the product by 7/2is 4/7,find the number.

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

4 0

Answer:

3/4

Step-by-step explanation:

$\frac{x}{y} * \frac{8}{3} * \frac{2}{7} = \frac{4}{7}\\\frac{3}{4} * \frac{8}{3} * \frac{2}{7} = \frac{4}{7 }$

You might be interested in

statuscvo [17]

Answer:

18ywjhtzk8ywbhuj

Step-by-step explanation:

ujzjtzjdjwjduxhxhwbsusuww

4 0

3 years ago

Jason is 4.52 feet tall his sister is 0.75 times his hight how tall is his sister

STatiana [176]

His sister is 7.895 feet tall.

3 0

4 years ago

First one to say "Free points" gets 50 points and also brainliest!

ycow [4]

Answer:

Free points...?

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The tax rate as a percent, r, charged on an item can be determined using the formula – 1 = r, where c is the final cost of the i

melisa1 [442]

The <em><u>correct answer</u></em> is:

$43.20

Explanation:

The formula we have is

c = p(1+r), where c is the total cost, p is the price of the item before tax, and r is the tax rate written as a decimal. This formula comes from the fact that adding a percent tax to the cost of an item takes 100% of the price and adds r% to it; this is why we multiply the price by (1+r).

Since our tax rate is 8%, r = 8% = 8/100 = 0.08. The price of the item is $40. Using this information, we have:

c = 40(1.08) = $43.20

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

3 years ago