If f(x) = 3x – 2 and g(x) = 6 – 4x, find f(x) + g(x)

Answer this volume based Question. I will make uh brainliest + 50 points

Harlamova29_29 [7]

Answer:

$\huge{\purple {r= 2\times\sqrt[3]3}}$

$\huge 2\times \sqrt [3]3 = 2.88$

Step-by-step explanation:

For solid iron sphere:
radius (r) = 2 cm (Given)

Formula for $V_{sphere}$ is given as:

$V_{sphere} =\frac{4}{3}\pi r^3$

$\implies V_{sphere} =\frac{4}{3}\pi (2)^3$

$\implies V_{sphere} =\frac{32}{3}\pi \:cm^3$

For cone:
r : h = 3 : 4 (Given)
Let r = 3x & h = 4x

Formula for $V_{cone}$ is given as:

$V_{cone} =\frac{1}{3}\pi r^2h$

$\implies V_{cone} =\frac{1}{3}\pi (3x)^2(4x)$

$\implies V_{cone} =\frac{1}{3}\pi (36x^3)$

$\implies V_{cone} =12\pi x^3\: cm^3$

It is given that: iron sphere is melted and recasted in a solid right circular cone of same volume
$\implies V_{cone} = V_{sphere}$

$\implies 12\cancel{\pi} x^3= \frac{32}{3}\cancel{\pi}$

$\implies 12x^3= \frac{32}{3}$

$\implies x^3= \frac{32}{36}$

$\implies x^3= \frac{8}{9}$

$\implies x= \sqrt[3]{\frac{8}{3^2}}$

$\implies x={\frac{2}{ \sqrt[3]{3^2}}}$

$\because r = 3x$

$\implies r=3\times {\frac{2}{ \sqrt[3]{3^2}}}$

$\implies r=3\times 2(3)^{-\frac{2}{3}}$

$\implies r= 2\times (3)^{1-\frac{2}{3}}$

$\implies r= 2\times (3)^{\frac{1}{3}}$

$\implies \huge{\purple {r= 2\times\sqrt[3]3}}$
Assuming log on both sides, we find:

$log r = log (2\times \sqrt [3]3)$

$log r = log (2\times 3^{\frac{1}{3}})$

$log r = log 2+ log 3^{\frac{1}{3}}$

$log r = log 2+ \frac{1}{3}log 3$

$log r = 0.4600704139$

Taking antilog on both sides, we find:

$antilog(log r )= antilog(0.4600704139)$

$\implies r = 2.8844991406$

$\implies \huge \red{r = 2.88\: cm}$

$\implies 2\times \sqrt [3]3 = 2.88$

8 0

2 years ago

Help me please I’m confused

Digiron [165]

Find ab….using your calculator

4 0

2 years ago

How do you do advanced order of operations? Need help on my homework!

Licemer1 [7]

Order of Operations = PEMDAS

Parentheses
Exponents
Multiplication
Division
Addition
Subtraction

6 0

3 years ago

F(x)= -x over x^2+5 on the domain [0,4]

BartSMP [9]

If you're asking for extrema, like the previous posting

well

$\bf f(x)=\cfrac{-x}{x^2+5}
\\\\\\
\cfrac{df}{dx}=\cfrac{(x^2+5)+2x^2}{(x^2+5)^2}\implies \cfrac{df}{dx}=\cfrac{5+3x^2}{(x^2+5)^2}\impliedby 
\begin{array}{llll}
using\ the\\
quotient\ rule
\end{array}$

like the previous posting, since this rational is identical, just that the denominator is negative, the denominator yields no critical points

and the numerator, yields no critical points either, so the only check you can do is for the endpoints, of 0 and 4

f(0) = 0 <---- only maximum, and thus absolute maximum

f(4) ≈ - 0.19 <---- only minimum, and thus absolute minimum

5 0

3 years ago

What must be true when solving a quadratic equation when using factoring?

Zigmanuir [339]

Answer:

Step-by-step explanation:

The function must be factorable.

For example, x^2 + x - 6= 0 factors to (x + 3)(x - 2) = 0 so the roots are -3 and 2.

x^2 + x - 7 = 0 will not factor so you need another method to solve this.

7 0

2 years ago