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2 years ago

|3x - 9| = 12 Show all steps please and thank you!

Mathematics

2 answers:

Brums [2.3K]2 years ago

6 0

Answer:

7 & -1

Step-by-step explanation:

Assuming you want the value of x...

Since the equation is wrapped in an absolute value, we must get x for both the positive and negative value.

Positive x

$3x-9=12\\3x=21\\x=7$

Negative x

$3x-9=-12\\3x=-3\\x=-1\\$

Both of these will equal 12 when tested.

$|3(-1)-9|=12\\|-3-9|=12\\|-12|=12\\12=12\\$

$|3(7)-9|=12\\|21-9|=12\\12=12\\$

bezimeni [28]2 years ago

4 0

Answer:

ayy I gotchuu

Step-by-step explanation:

theres gonna be two solutions

a negative then positive

positive:

|3x - 9| = 12

3x-9=12

3x=21

x=7

negative:

3x-9=-12

3x=-3

x=-1

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Adam buys a box of stuff that weighed 6 3/8. if he buys box of fruit that weighed 7 2/5. what is the combined weight.

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Answer:

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Step-by-step explanation:

Given:

Adam buys a box of stuff that weighed 6 3/8. if he buys box of fruit that weighed 7 2/5.

Now, to find the combined weight.

So, we change the mixed fraction to improper first:

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Then, by adding to get the combined weight:

$\frac{51}{8}+\frac{37}{5}$

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

kirill115 [55]

Answer:

Required largest volume is 0.407114 unit.

Step-by-step explanation:

Given surface area of a right circular cone of radious r and height h is,

$S=\pi r\sqrt{r^2+h^2}$

and volume,

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

To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

$f(r,h)=\frac{1}{3}\pi r^2 h$

subject to,

$g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)$

We know for maximum volume $r\neq 0$ . So let $\lambda$ be the Lagranges multipliers be such that,

$f_r=\lambda g_r$

$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

And,

$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

$\implies \pi r \sqrt{2r^2+r^2}=8$

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Then,

$h=\sqrt{2}(1.21252)\equiv 1.71476$

Hence largest volume,

$V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114$

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