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3 years ago

What is the last one number 12

Mathematics

1 answer:

rosijanka [135]3 years ago

8 0

What is the last one number 12

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Solve this function f(-4)= -3x+6

timama [110]

Answer:

Solve for f.

f(−4)=−3x+6

Step 1: Divide both sides by -4.

−4f/-4 = −3x+6/−4

f=3/4x + −3/2

Answer:

f= 3/4x + −3/2

7 0

4 years ago

The product of 2 and the sum of a numbet and 8 is equivalent to 16

Rudik [331]

2(x+8)=16

2x+16=16

2x=16-16

2x=0

x=0

6 0

3 years ago

Factor completely <br> 4(x+1)^2/3 + 12(x+1)^-1/3

wolverine [178]

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
\left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}
\\\\
-------------------------------\\\\
%4(x+1)^2/3 + 12(x+1)^-1/3
4(x+1)^{\frac{2}{3}}+12(x+1)^{-\frac{1}{3}}\implies 4(x+1)^{\frac{2}{3}}+12\cfrac{1}{(x+1)^{\frac{1}{3}}}
\\\\\\$

$\bf 4(x+1)^{\frac{2}{3}}+\cfrac{12}{(x+1)^{\frac{1}{3}}}\impliedby \textit{so, our LCD is }(x+1)^{\frac{1}{3}}
\\\\\\
\cfrac{4(x+1)^{\frac{2}{3}}\cdot (x+1)^{\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)^{\frac{2}{3}+\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4(x+1)^{\frac{3}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4x+4+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4x+16}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+4)}{\sqrt[3]{x+1}}$

3 0

4 years ago

Read 2 more answers

:(NOBODY HELPS ME PLZZ HELP ME

Neko [114]

Hello!

$\large\boxed{B.}$

Find the slope using the slope formula:

m = (y2 - y1) / (x2 - x1)

Plug in the points (-2, 8) and (0, 6) to solve:

m = (6 - 8) / (0 - (-2))

m = -2 / 2

m = -1

The y-intercept of the function is the y-value when x = 0.

According to the table, at x = 0, y = 6. Therefore, the y-intercept is (0, 6).

The correct answer is B.

4 0

3 years ago

Express each cubic function below in terms of linear factors.a)f(x)=x³-6x²-27xb)f(x)=x³-16x²c)f(x)=x³+16x²

nirvana33 [79]

Answer:

a) x (x-9) (x+3)

b) x (x+4)(x-4)

c) (x)(x)(x + 16)

Step-by-step explanation:

We're going to equal the functions to zero and find their roots by factorizing:

a) f(x)=x³-6x²-27x

x³-6x²-27x = 0

x(x²-6x-27) = 0

⇒x (x-9) (x+3)

b) f(x)=x³-16x

x³-16x = 0

x (x²-16) =0

⇒x (x+4)(x-4)

c) f(x)=x³+16x²

x³ + 16x² = 0

x² ( x +16)

⇒(x)(x)(x + 16)

3 0

4 years ago