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2 years ago

Find the product of the binomials, (x - 3)(x + 2) (x + 4) =

Mathematics

2 answers:

MrRa [10]2 years ago

7 0

Answer:

x^3+3x^2-10x-24

Step-by-step explanation:

(x - 3)(x + 2) (x + 4)

multiply the first two together first

(x^2+2x-3x-6)(x+4) add like terms first

(x^2-x-6)(x+4)

then multiply

x^3+4x^2-x^2-4x-6x-24

add like terms

x^3+3x^2-10x-24

erastovalidia [21]2 years ago

6 0

Answer:

Step-by-step explanation:

(x - 3)(x + 2) (x + 4) =

x^3+3x^2-10x-24

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Please help me with this question.

Korvikt [17]

2x is what I would say but i sayy ask an expert

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3 years ago

The perimeter of a rectangle is 98 m. The length of the rectangle is 9 m more than four times the width. Find the dimensions of

Olenka [21]

P=98 m

l +9=4w

p=2l+2w

98=2(4w-9) +2w

98 = 8w-18+2w

116=10w

w=11,6 m

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l= 37,4 m

3 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

The pair of shoes you want is on sale for 25% off. If the shoes originally cost $65, how much is the sale price

Ludmilka [50]

$48.75
Good this help. Ask if you want me to explain :)

5 0

3 years ago

Read 2 more answers

Solve the following system of equations by elimination:

sergeinik [125]

Answer:

x=4

y=5

(4,5)

Step-by-step explanation:

6 0

2 years ago