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3 years ago

Please show your work this question what made you come to the conclusion. Thank you

Mathematics

1 answer:

AveGali [126]3 years ago

8 0

Answer:

B the range, the x- and y-intercept

Step-by-step explanation:

the domain stays the same : all values of x are possible out of the interval (-infinity, +infinity).

but the range changes, as for the original function y could only have positive values - even for negative x.

the new function has a first term (with b) that can get very small for negative x, and then a subtraction of 2 makes the result negative.

the y-intercept (x=0) of the original function is simply y=1, as b⁰=1.

the y-intercept of the new function is definitely different, because the first term 3×(b¹) is larger than 3, because b is larger than 1. and a subtraction of 2 leads to a result larger than 1, which is different to 1.

the original function has no x-intercept (y=0), as this would happen only for x = -infinity. and that is not a valid value.

the new function has an x-intercept, because the y-values (range) go from negative to positive numbers. any continuous function like this must therefore have an x-intercept (again, y = the function result = 0)

$3 {b}^{x + 1} = 2$

${b}^{x + 1} = 2 \div 3$

$log_{b}(2 \div 3) = x + 1$

$x = log_{b}(2 \div 3) - 1$

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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,

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

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

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

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

HELP QUICK LOL THANKS !!!

Anon25 [30]

Answer:

0.6

Step-by-step explanation:

The x increases by 5, while y increases by 3

Lets use the method, which is $\frac{rise}{run}$ .

Keep in mind that x goes horizontally along the axis, while y goes vertically on the axis.

So plug the numbers in.

$\frac{3}{5} =$ 0.6

Hence, the constant proportionality is 0.6.

8 0

3 years ago

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Least to greatest !?

Bad White [126]

The lengths are $\frac{5}{6} \frac{5}{12} and \frac{1}{3}$
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3 years ago

Need answers for 3. A & B and 4

Ivan

I think to N because it make more sense in the number line

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3 years ago

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Umber 8. What is the value of the expression - -g-(-5) =-9 - 5 5

pantera1 [17]

First, we want to subtract the numbers within the parenthesis. However, to do this we must put each number over a common denominator. We can do his by multiplying

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]

−

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(

−

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(

)

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