Solve this inequality: j/4 – 8 < 4. A. j < –48 B. j < 48 C. j > –12 D. j < 12

Mathematics

2 answers:

postnew [5]3 years ago

4 0

J/4 - 8 < 4

j/4 < 4 + 8
j4 < 12
j < 12 X 4
j < 48

Kruka [31]3 years ago

3 0

$\frac{j}{4} - 8 \ \textless \ 4 \\ \\ \frac{j}{4} \ \textless \ 4 + 8 \\ \\ \frac{j}{4} \ \textless \ 12 \\ \\ j \ \textless \ 12 \times 4 \\ \\ j \ \textless \ 48 \\ \\$

The final result is: j < 48.

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. A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius of the base and height) of the drum

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Answer:

$r=1.05\ \text{ft}$

$h=2.12\ \text{ft}$

$20.91\ \text{ft}^3$

Step-by-step explanation:

r = Radius

h = Height

Volume of cylinder = $7.35\ \text{ft}^3$

$V=\pi r^2h\\\Rightarrow h=\dfrac{V}{\pi r^2}\\\Rightarrow h=\dfrac{7.35}{\pi r^2}$

Surface area is given by

$A=2\pi r^2+2\pi rh\\\Rightarrow A=2\pi r^2+2\pi r\dfrac{7.35}{\pi r^2}\\\Rightarrow A=2\pi r^2+\dfrac{14.7}{r}$

Differentiating with respect to radius we get

$\dfrac{dA}{dr}=4\pi r-\dfrac{14.7}{r^2}$

Equating with zero we get

$0=4\pi r-\dfrac{14.7}{r^2}\\\Rightarrow 4\pi r=\dfrac{14.7}{r^2}\\\Rightarrow r^3=\dfrac{14.7}{4\pi}\\\Rightarrow r=(\dfrac{14.7}{4\pi})^{\dfrac{1}{3}}\\\Rightarrow r=1.05$

$\dfrac{d^2A}{dr^2}=4\pi-2\times \dfrac{14.7}{r^3}\\\Rightarrow \dfrac{d^2A}{dr^2}=4\pi+2\times \dfrac{14.7}{1.05^3}\\\Rightarrow \dfrac{d^2A}{dr^2}=37.96>0$