Help help help if u can answer this I’ll mark u brainliest (6) (the one in the middle)

Mathematics

1 answer:

Hoochie [10]4 years ago

4 0

You just have to plug in each thing:

So f(0):

3(0)^2 - 4 = -4

so its not A

f(-2) and f(2):

3(-2)^2 - 4

3(2)^2 -4

now we don't even have to calculate these, anything to the power of an even number is a positive number so -2^2 and 2^2 both equal 4

B is correct, so you can do the others if you want to check, but if B is true the others shouldn't be true.

You might be interested in

What are the factors of 12x^3y^3--24x^2y^3 Factor completely and show work.

Fed [463]

Already factored:

Factors of 12X^3y^3 - 24x^2y^3

Answer:

The Factors of 12 are: 1, 2, 3, 4, 6, 12

The factor pairs of 12 are:
1(12) = 12
2(6) = 12
3(4) = 12

Hope this helps.!!!

5 0

3 years ago

Read 2 more answers

Plz help I have no idea how to answer it

saveliy_v [14]

5. Answer: see explanation

Step-by-step explanation:

If the roots are m and 3m, then x = m and x = 3m

⇒ x - m = 0 and x - 3m = 0

⇒ (x - m)(x - 3m) = 0

⇒ x² - 4mx + 3m² = 0

Since x² + px + q = 0 then p = -4m and q = 3m²

3p² = 3(-4m)² 16q = 16(3m²)

= 3(16m²) = 48m²

= 48m²

3p² = 48m² = 16q ⇒ 3p² = 16q

**********************************************************************************************

6. Answer: 8 or 18

Step-by-step explanation:

The Area of the entire rectangle (A = L × w) is 12 × 10 = 120

The Area of the shaded region is 72, so the Area of the non-shaded region is 120 - 72 = 48.

There are two non-shaded triangles.

Bigger non-shaded Δ: L = 12-x, w = 10 ⇒ $A = \dfrac{10(12-x)}{2}$
Smaller non-shaded Δ: L = x, w = x ⇒ $A=\dfrac{x(x)}{2}$

Combine the Areas of both triangles and set it equal to the Area of the non-shaded region:

$\dfrac{x(x)}{2}+\dfrac{10(12-x)}{2}=48\\\\\\\dfrac{x^2}{2}+\dfrac{120-10x}{2}=48\\\\\\\dfrac{x^2-10x+120}{2}=48\\\\\\x^2-10x+120=96\\\\x^2-10x+24=0\\\\(x-4)(x-6)=0\\\\x - 4 = 0\quad or\quad x-6=0\\x=4\qquad or\qquad x=6\\\\\text{Based on the image provided, it appears that the x-value will be the}\\\text{smallest of the two possible solutions (4), but both solutions are valid.}$