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3 years ago

Multiply the following and simplify, show all work (3x^2+6x−6)(5x^2−3)

2 answers:

8 0

This has the same rules as double brackets, except you have to multiply 3 terms by 2 terms. First you multiply them all by 5x^2:
3x^2 × 5x^2 = 15x^4
6x × 5x^2 = 30x^3
-6 × 5x^2 = -30x^2
Now by -3:
3x^2 × -3 = -9x^2
6x × -3 = -18x
-6 × -3 = 18

So then you add up the like terms and you get 15x^4 + 30x^3 - 39x^2 - 18x + 18, which is as far as you can simplify. I hope this helps!

AnnyKZ [126]3 years ago

7 0

(3x^2+6x−6)(5x^2−3)
=15x^4 - 9x^2 + 30x^3 - 18x - 30x^2 + 18
=15x^4 + 30x^3 - 39x^2 - 18x + 18

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Carl and Derrick are part-time employees. They are paid different hourly rates. One week, Carl worked for 15 hours and Derrick w

Bingel [31]

Answer:

Carl's hourly pay rate is $16

Step-by-step explanation:

Let x is the hourly rate of Carl and y is the hourly rate of Derrick

∵ Carl's hourly rate is $x

∵ Derrick's hourly rate is $y

∵ Carl worked for 15 hours and Derrick worked for 20 hours

∴ Their combined pay was 15x + 20y

∵ Their combined pay was $640

→ Equate the value of the combined pay

∴ 15x + 20y = 640 ⇒ (1)

∵ The next week, Carl worked 20 hours and Derrick worked 25 hours

∴ Their combined pay was 20x + 25 y

∵ Their combined pay was $820

→ Equate the value of the combined pay

∴ 20x + 25y = 820 ⇒ (2)

Now we have a system of equations to solve it

→ Multiply equation (1) by 5 and equation (2) by -4

∵ 5(15x) + 5(20y) = 5(640)

∴ 75x + 100y = 3200 ⇒ (3)

∵ -4(20x) + -4(25y) = -4(820)

∴ -80x - 100y = -3280 ⇒ (4)

→ Add equations (3) and (4)

∵ (75x + -80x) + (100y + -100y) = (3200 + -3280)

∴ -5x + 0 = -80

∴ -5x = -80

→ Divide both sides by -5 to find x

∴ x = 16

∴ Carl's hourly pay rate is $16

7 0

3 years ago

Read 2 more answers

When a normal distribution is represented by a histogram, where would most scores fall?

Doss [256]

Answer:

Most scores will fall right in the middle of the graph where the mean score lies.

Step-by-step explanation:

We are given that a distribution is normal.

The distribution is represented by a histogram, thus, the histogram would have a bell shape.

Also, since the distribution is normal, the mean, mode , median of the distribution is same.

Thus, we could say that the modal value for the normal distribution is the mean value.

Thus, most scores will fall right in the middle of the graph where the mean score lies.

4 0

4 years ago

Given that angle a= 168 work out x REALLY NEED HELP 21P !!

Zina [86]

One angle = 90°
the other 3 angles = 360 - 90 = 270

so
x + 2x + a = 270 but a = 168
3x + 168 = 270
3x = 102
x = 102/3
x = 34

answer
x = 34

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3 years ago

Based on the given inequality statement, which statement correctly describes where the values are located on a number line rela

KonstantinChe [14]

Answer:

what are the different choices?

Step-by-step explanation:

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3 years ago

Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE

inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

we have

$A(5,0), B(2,4), C(-2,1),D(1,-3)$

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(5,0),B(2,4)$

substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

$d=\sqrt{(4)^{2}+(-3)^{2}}$

$d=\sqrt{25}$

$AB=5\ units$

Find the distance BC

$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$BC=5\ units$

Find the distance CD

$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$CD=5\ units$

Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

$B(2,4),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

$d=\sqrt{(-7)^{2}+(-1)^{2}}$

$BD=\sqrt{50}\ units$

Verify if the polygon is a square

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

The triangle BDA is a right triangle

therefore

The polygon is a square

Find the Area of the polygon

The area of a square is equal to

$A=b^{2}$

we have

$b=5\ units$

$A=5^{2}=25\ units^{2}$

Find the perimeter of the polygon

The perimeter of a square is equal to

$P=4b$

we have

$b=5\ units$

$P=4(5)=20\ units$

6 0

3 years ago