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3 years ago

The polynomial 4x^3 - 6x^2 + 8x – 12 can be grouped in different ways to factor by

Mathematics

1 answer:

zubka84 [21]3 years ago

8 0

Answer:

(4x3-6x) + (8x-12)= 2x

(4x3+8x) + (-6x2-12)= 2(2x-3) (x^2+2)

Step-by-step explanation:

All you have to do is simplify.

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Help meeee plz I don’t get it

MatroZZZ [7]

Answer: F

Step-by-step explanation:

This is basically like 2 rectangles. Find the area of both and add them together.

Remember that the area is $l$ x $w$ .

Area of Rectangle 1: $l$ = 7, $w$ = 14

14 x 7 = 98 $cm^{2}$

Area of Rectangle 2: $l$ = 5, $w$ = 14

14 x 5 = 70 $cm^{2}$

Add the areas together: 98 $cm^{2}$ + 70 $cm^{2}$ = 168 $cm^{2}$

6 0

2 years ago

A bottle of perfume holds 0.55 onice. A bottle of cologne holds p.2 ounce. How many more ounces dose the bottle of perfume hold?

zimovet [89]

It has .33 more ounces if that says .2

7 0

2 years ago

8+8- 19=??<br> pls help and thank u

kumpel [21]

-3 the answer is -3 ight

4 0

3 years ago

Read 2 more answers

Sin x= 1.5<br> value of x

alina1380 [7]

Answer:

x = π - sin^(-1)(3/2) + 2 π n_1 for n_1 element Z

or x = 2 π n_2 + sin^(-1)(3/2) for n_2 element Z

Step-by-step explanation:

Solve for x:

sin(x) = 1.5

1.5 = 3/2:

sin(x) = 3/2

Take the inverse sine of both sides:

Answer: x = π - sin^(-1)(3/2) + 2 π n_1 for n_1 element Z

or x = 2 π n_2 + sin^(-1)(3/2) for n_2 element Z

3 0

3 years ago

A manufacturer produces bolts of a fabric with a fixed width. A quantity q of this fabric (measured in yards) that is sold is a

BigorU [14]

Answer:

$R'(20) = 2000$

Step-by-step explanation:

We are given the following in the question:

Quantity, q

Selling price in dollars per yard, p

$q=f(p)$

Total revenue earned =

$R(p)=pf(p)$

f(20)=13000

This means that 13000 yards of fabric is sold when the selling price is 20 dollars per yard.

f′(20)=−550

This means that increasing the selling price by 1 dollar per yards there is a decrease in fabric sales by 550.

We have to find R'(20)

Differentiating the above expression, we have,

$R'(p) = \displaystyle\frac{d(R(p))}{dp} = \frac{d(pf(p))}{dp} = f(p) + pf'(p)$

Putting the values, we get,

$R'(p) = f(p) + pf'(p)\\\\R'(20) = f(20) + 20(f'(20))\\\\R'(20) = 13000 + 20(-550) = 2000$

7 0

3 years ago