Ask question

Ask question

All categories

3 years ago

7

If f(x) = -4x^2 - 6x - 1 and g(x) = -x^2 - 5x + 3, find (f + g)(x).

1 answer:

Flauer [41]3 years ago

7 0

Answer:

-5 $x^{2}$ -11x+2

Step-by-step explanation:

-4 $x^{2} \\$ -6x-1- $x^{2}$ -5x+3=-5 $x^{2}$ -11x+2

You might be interested in

What is the molar solubility of magnesium fluoride in a solution that is 0.40 m f− ions? the ksp of mgf2 is 6.4 × 10−9. hints wh

velikii [3]

<span>Answer: MgF2 ====â‡’ Mg+2 + 2F- Ksp = [Mg+2][F-]^2 Let S = molar solubility of MgF2 [Mg+2] = S ; [F-] = 2s Since the [F-] is initially 0.40 M, then [F-] = 0.40 + 2S 6.4 x 10^-9 = (S) (0.40 + 2S)^2 ; one can neglect the 2S in the 0.40 + 2S expression since it is very, very small compared to the 0.40 already present. 6.4 x 10^-9 = S(0.40)^2 S = 4.0 x 10^-8 Molar solubility = 4.0 x 10^-8</span>

6 0

3 years ago

Plz help its 7th grade dis trubtive proprety

alina1380 [7]

It would be -4x+(-12)=12 good luck

8 0

3 years ago

Read 2 more answers

#3 AB and AD are tangent to the circle centered at point C. Find the value of x. *

dezoksy [38]

Answer:

x=4

Step-by-step explanation:

tangent lines from the same point to a circle are congruent in length, so we can say that

5x+8 = 8x-4

5x -5x +8 = 8x - 5x -4

8 = 3x - 4

8+4 = 3x -4 + 4

3x = 12

3x/3 = 12/3

x=4

8 0

3 years ago

Answer please no links or files

Naddik [55]

Answer:

120

Step-by-step explanation:

4 0

3 years ago

A triangle has base 2x+1 and height 6x-3. What value of x would give an area of 240m2?

GREYUIT [131]

Answer:

The values of x which would give an area of 240m² would be:

$x=-\frac{\sqrt{161}}{2},\:x=\frac{\sqrt{161}}{2}$

Step-by-step explanation:

Given

The base of triangle b = 2x+1

The height of triangle h = 6x-3

The Area of the triangle A = 240 m²

The Area of the triangle has the formula

A = 1/2 × b × h

substituting b = 2x+1, h = 6x-3 and A = 240

$240\:=\:\frac{1}{2}\:\left(2x+1\right)\:\times \left(6x-3\right)$

$480=\left(2x+1\right)\left(6x-3\right)$

$480=12x^2-3$

Subtract 480 from both sides

$12x^2-3-480=480-480$

$12x^2-483=0$

$3\left(4x^2-161\right)=0$

$3\left(2x+\sqrt{161}\right)\left(2x-\sqrt{161}\right)=0$

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

$2x+\sqrt{161}=0\quad \mathrm{or}\quad \:2x-\sqrt{161}=0$

solving

$2x+\sqrt{161}=0$

$2x=-\sqrt{161}$

Divide both sides by 2

$\frac{2x}{2}=\frac{-\sqrt{161}}{2}$

$x=-\frac{\sqrt{161}}{2}$

also solving

$2x-\sqrt{161}=0$

$2x=\sqrt{161}$

Divide both sides by 2

$\frac{2x}{2}=\frac{\sqrt{161}}{2}$

$x=\frac{\sqrt{161}}{2}$

Therefore, the values of x which would give an area of 240m² would be:

$x=-\frac{\sqrt{161}}{2},\:x=\frac{\sqrt{161}}{2}$

3 0

2 years ago