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2 years ago

a computer and printer a totoal cost $1132 the cost of the computer is three times the cost of the ptinter

1 answer:

3 0

Use a system of equations

C+P=1132
3P=C

Substitute C in first equation as
3P+P=1132
Simplify
4P=1132
Solve
P=1132/4
P=283

NOW SOLVE FOR C SUBSTITUTING P VALUE IN FIRST EQUATION

C+283=1132
C=1132-283
C=849

Printer = 283$
Computer = 849$

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What is the time and space margin that is the distance you can see ahead of your vehicle?

IgorLugansk [536]

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Insufficient information is provided to be more specific.

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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

Which real-world scenario can be described using the algebraic expression c/4?

Romashka [77]

Answer:

Baking c cookies and dividing them evenly among 4 friends.

Step-by-step explanation:

7 0

3 years ago

Steven has a rectangular rug with a perimeter of 16 feet. The width of the rug is 5 feet. What is the length of the rug?

boyakko [2]

$\boxed{\boxed{\text{Perimeter = 2(Length + Width)}}}$

Given that the perimeter is 16ft and width is 5 ft,
$16 = 2( Length + 5)$

Distribute 2:
$16 = 2Length + 10$

Take away 10 from both sides:
$2Length = 16 - 10$

Evaluate the right hand side:
$2Length = 6$

Divide by 2 on both sides:
$Length = 3$

$\boxed{\boxed{\text{Answer: Length = 3 ft}}}$

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3 years ago

Use substitution to solve this system:

Oliga [24]

awnser:

y = 62 /3

Step-by-step explanation:

5(6)+3y=92

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3 years ago