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yKpoI14uk [10]
3 years ago
12

What is the product? (-2x- 9y2 )(-4x-3)

Mathematics
1 answer:
lara [203]3 years ago
4 0

Answer:

36xy^2 + 8x^2 + 27y^2 + 6x

Step-by-step explanation:

Use the FOIL method of multiplying binomials.

First term in each binomial: -2x * -4x = 8x^2

Outside terms: -2x * -3 = 6x

Inside terms: -9y^2 * -4x = 36xy^2

Last term in each binomial: -9y^2 * -3 = 27y^2

Now, rearrange the terms correctly. 36xy^2 + 8x^2 + 27y^2 + 6x

This is our final answer, since it can not be simplified any more.

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Evaluate the following expression: (show work)
Scilla [17]
3(3)^2 - 4^3 - 4^3 - (-5)
3(9) - 64 - 64 + 5
27 - 64 - 64 + 5
-96
4 0
3 years ago
Determine whether the points (–3,–2) and (3,2) are in the solution set of the system of inequalities below. y ≤ ½x + 2 y < –2
lana [24]

Given:

The system of inequalities:

y\leq \dfrac{1}{2}x+2

y

To find:

Whether the points (–3,–2) and (3,2) are in the solution set of the given system of inequalities.

Solution:

A point is in the solution set of the given system of inequalities if it satisfies both inequalities.

Check for the point (-3,-2).

-2\leq \dfrac{1}{2}(-3)+2

-2\leq -1.5+2

-2\leq 0.5

This statement is true.

-2

-2

-2

This statement is also true.

Since the point (-3,-2) satisfies both inequalities, therefore (-3,-2) is in the solution set of the given system of inequalities.

Now, check for the point (3,2).

2

2

2

This statement is false because 2>-9.

Since the point (3,2) does not satisfy the second inequality, therefore (3,2) is not in the solution set of the given system of inequalities.

7 0
3 years ago
Factor completely: (4c - x)^2 - (2c + 3x)^2
azamat

Answer:

Graph for (4*​2.99792e+08-​x)^​2-​(2*​2.99792e+08+​3*​x)^​2

More info

How to solve your problem

(4−)2−1(2+3)2

(4c−x)2−1(2c+3x)2(4c-x)^{2}-1(2c+3x)^{2}(4c−x)2−1(2c+3x)2

Simplify

1

Expand the square

(4−)2−1(2+3)2

(4c−x)2−1(2c+3x)2\left(4c-x\right)^{2}-1(2c+3x)^{2}(4c−x)2−1(2c+3x)2

(4−)(4−)−1(2+3)2

(4c−x)(4c−x)−1(2c+3x)2(4c-x)(4c-x)-1(2c+3x)^{2}(4c−x)(4c−x)−1(2c+3x)2

2

Distribute

(4−)(4−)−1(2+3)2

(4c−x)(4c−x)−1(2c+3x)2{\color{#c92786}{(4c-x)(4c-x)}}-1(2c+3x)^{2}(4c−x)(4c−x)−1(2c+3x)2

4(4−)−(4−)−1(2+3)2

4c(4c−x)−x(4c−x)−1(2c+3x)2{\color{#c92786}{4c(4c-x)-x(4c-x)}}-1(2c+3x)^{2}4c(4c−x)−x(4c−x)−1(2c+3x)2

3

Distribute

4(4−)−(4−)−1(2+3)2

4c(4c−x)−x(4c−x)−1(2c+3x)2{\color{#c92786}{4c(4c-x)}}-x(4c-x)-1(2c+3x)^{2}4c(4c−x)−x(4c−x)−1(2c+3x)2

162−4−(4−)−1(2+3)2

16c2−4cx−x(4c−x)−1(2c+3x)2{\color{#c92786}{16c^{2}-4cx}}-x(4c-x)-1(2c+3x)^{2}16c2−4cx−x(4c−x)−1(2c+3x)2

4

Distribute

162−4−(4−)−1(2+3)2

16c2−4cx−x(4c−x)−1(2c+3x)216c^{2}-4cx{\color{#c92786}{-x(4c-x)}}-1(2c+3x)^{2}16c2−4cx−x(4c−x)−1(2c+3x)2

162−4−4+2−1(2+3)2

16c2−4cx−4cx+x2−1(2c+3x)216c^{2}-4cx{\color{#c92786}{-4cx+x^{2}}}-1(2c+3x)^{2}16c2−4cx−4cx+x2−1(2c+3x)2

5

Combine like terms

162−4−4+2−1(2+3)2

16c2−4cx−4cx+x2−1(2c+3x)216c^{2}{\color{#c92786}{-4cx}}{\color{#c92786}{-4cx}}+x^{2}-1(2c+3x)^{2}16c2−4cx−4cx+x2−1(2c+3x)2

162−8+2−1(2+3)2

16c2−8cx+x2−1(2c+3x)216c^{2}{\color{#c92786}{-8cx}}+x^{2}-1(2c+3x)^{2}16c2−8cx+x2−1(2c+3x)2

6

Expand the square

162−8+2−1(2+3)2

16c2−8cx+x2−1(2c+3x)216c^{2}-8cx+x^{2}-1\left(2c+3x\right)^{2}16c2−8cx+x2−1(2c+3x)2

162−8+2−1(2+3)(2+3)

16c2−8cx+x2−1(2c+3x)(2c+3x)16c^{2}-8cx+x^{2}-1(2c+3x)(2c+3x)16c2−8cx+x2−1(2c+3x)(2c+3x)

7

Distribute

162−8+2−1(2+3)(2+3)

16c2−8cx+x2−1(2c+3x)(2c+3x)16c^{2}-8cx+x^{2}-1{\color{#c92786}{(2c+3x)(2c+3x)}}16c2−8cx+x2−1(2c+3x)(2c+3x)

162−8+2−1(2(2+3)+3(2+3))

16c2−8cx+x2−1(2c(2c+3x)+3x(2c+3x))16c^{2}-8cx+x^{2}-1({\color{#c92786}{2c(2c+3x)+3x(2c+3x)}})16c2−8cx+x2−1(2c(2c+3x)+3x(2c+3x))

8

Distribute

162−8+2−1(2(2+3)+3(2+3))

16c2−8cx+x2−1(2c(2c+3x)+3x(2c+3x))16c^{2}-8cx+x^{2}-1({\color{#c92786}{2c(2c+3x)}}+3x(2c+3x))16c2−8cx+x2−1(2c(2c+3x)+3x(2c+3x))

162−8+2−1(42+6+3(2+3))

16c2−8cx+x2−1(4c2+6cx+3x(2c+3x))16c^{2}-8cx+x^{2}-1({\color{#c92786}{4c^{2}+6cx}}+3x(2c+3x))16c2−8cx+x2−1(4c2+6cx+3x(2c+3x))

9

Distribute

162−8+2−1(42+6+3(2+3))

16c2−8cx+x2−1(4c2+6cx+3x(2c+3x))16c^{2}-8cx+x^{2}-1(4c^{2}+6cx+{\color{#c92786}{3x(2c+3x)}})16c2−8cx+x2−1(4c2+6cx+3x(2c+3x))

162−8+2−1(42+6+6+92)

16c2−8cx+x2−1(4c2+6cx+6cx+9x2)16c^{2}-8cx+x^{2}-1(4c^{2}+6cx+{\color{#c92786}{6cx+9x^{2}}})16c2−8cx+x2−1(4c2+6cx+6cx+9x2)

10

Combine like terms

162−8+2−1(42+6+6+92)

16c2−8cx+x2−1(4c2+6cx+6cx+9x2)16c^{2}-8cx+x^{2}-1(4c^{2}+{\color{#c92786}{6cx}}+{\color{#c92786}{6cx}}+9x^{2})16c2−8cx+x2−1(4c2+6cx+6cx+9x2)

162−8+2−1(42+12+92)

16c2−8cx+x2−1(4c2+12cx+9x2)16c^{2}-8cx+x^{2}-1(4c^{2}+{\color{#c92786}{12cx}}+9x^{2})16c2−8cx+x2−1(4c2+12cx+9x2)

11

Distribute

162−8+2−1(42+12+92)

16c2−8cx+x2−1(4c2+12cx+9x2)16c^{2}-8cx+x^{2}{\color{#c92786}{-1(4c^{2}+12cx+9x^{2})}}16c2−8cx+x2−1(4c2+12cx+9x2)

162−8+2−42−12−92

16c2−8cx+x2−4c2−12cx−9x216c^{2}-8cx+x^{2}{\color{#c92786}{-4c^{2}-12cx-9x^{2}}}16c2−8cx+x2−4c2−12cx−9x2

12

Combine like terms

162−8+2−42−12−92

16c2−8cx+x2−4c2−12cx−9x2{\color{#c92786}{16c^{2}}}-8cx+x^{2}{\color{#c92786}{-4c^{2}}}-12cx-9x^{2}16c2−8cx+x2−4c2−12cx−9x2

122−8+2−12−92

12c2−8cx+x2−12cx−9x2{\color{#c92786}{12c^{2}}}-8cx+x^{2}-12cx-9x^{2}12c2−8cx+x2−12cx−9x2

13

Combine like terms

122−8+2−12−92

12c2−8cx+x2−12cx−9x212c^{2}{\color{#c92786}{-8cx}}+x^{2}{\color{#c92786}{-12cx}}-9x^{2}12c2−8cx+x2−12cx−9x2

122−20+2−92

12c2−20cx+x2−9x212c^{2}{\color{#c92786}{-20cx}}+x^{2}-9x^{2}12c2−20cx+x2−9x2

14

Combine like terms

122−20+2−92

12c2−20cx+x2−9x212c^{2}-20cx+{\color{#c92786}{x^{2}}}{\color{#c92786}{-9x^{2}}}12c2−20cx+x2−9x2

122−20−82

Step-by-step explanation:

7 0
3 years ago
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Taya2010 [7]

Answer:

30seconds

Step-by-step explanation:

if it takes 2 seconds to walk 3steps then in 30 seconds she should have walked 60 steps.

3 0
3 years ago
A circle is shown. Points Q, U, A, D are on the circle. Lines connect the points to form a quadrilateral. Angle Q U A is 111 deg
PilotLPTM [1.2K]

Answer:

b. 50

Step-by-step explanation:

6 0
4 years ago
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