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tekilochka [14]

3 years ago

8

One less than the quotient of four and number x

1 answer:

hichkok12 [17]3 years ago

8 0

4/x-1 would be the equation written from that statement.

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Use the probability distribution graph to answer the question.

vivado [14]

Answer:

5

Step-by-step explanation:

So we want to find the value of x where the area to the left of it is equal to 0.6.

Let's start by finding the area of the first triangle, between x=0 and x=4.

A = 1/2 bh

A = 1/2 (4) (0.2)

A = 0.4

So we know a > 4. What if we add the area of that rectangle?

A = 0.4 + bh

A = 0.4 + (1) (0.2)

A = 0.6

Aha! So a = 5.

7 0

3 years ago

HELP!!! Which of the following is not a perfect square trinomial? A. 169 – 26y + y2 B. 81 + 18y + y2 C. 64 + 8y + y2 D. 25 + 10y

natita [175]

Let's review the options:
A. 169-26y+y^2 can be factorised into (13-y)(13-y)
B. 81+18y+y^2 can be factorised into (9+y)(9+y)
C. 64+8y+y^2 cannot be factorised into double brackets
D. 25+10y+y^2 can be factorised into (5+y)(5+y)

Therefore your answer is the odd one out, C.

3 0

3 years ago

7. Write the equation of the line that is perpendicular to y = -5x + 1 and passes through the point (2,-1). slope intercept equa

Svetach [21]

Answer:

y = 1/5x - 7/5

Step-by-step explanation:

y = 1/5x + b

-1 = 1/5(2) + b

-1 = 2/5 + b

-7/5 = b

4 0

2 years ago

4. 4c - 3 (c = -2)<br><br><br> will give 5 stars

Flauer [41]

4(-2)-3 = -8-3 = -11
Answer: -11

4 0

3 years ago

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,

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

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

$\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252$

Then,

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

Hence largest volume,

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

3 0

3 years ago