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3 years ago

11

Six times the sum of a number and twelve is forty.

1 answer:

solniwko [45]3 years ago

4 0

X-the number

6(x + 12) = 40 |use distributive property: a(b + c) = ab + ac

6x + 72 = 40 |subtract 72 from both sides

6x = -32 |divide both sides by 6

x = -32/6

x = -16/3

x = -5 1/3

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Which statements describe the graph of y = 3√x-1+2 ? Check all that apply.

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Answer:

The correct options are A, D and E.

Step-by-step explanation:

We are given the function $y=\sqrt[3]{x-1} +2$

From the graph below, we see that,

Domain and range of the function is the set of all real numbers.

<em>As y-intercept is the point where the graph of the function crosses y-axis.</em>

That is, y-intercept is obtained when x= 0.

So, on substituting, we have,

$y=\sqrt[3]{0-1} +2$

i.e. $y=-1+2$

i.e. y= 1

Thus, the y-intercept is (0,1).

<em>Also, x-intercept is the point where the graph of the function crosses x-axis.</em>

That is, x-intercept is obtained when y= 0.

So, on substituting, we have,

$0=\sqrt[3]{x-1} +2$

i.e. $-2=\sqrt[3]{x-1}$

i.e. $-8=x-1$

i.e. x= -7

Thus, the x-intercept is (-7,0).

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3 years ago

Pam has 90 m of fencing to enclose an area in a petting zoo with two dividers to separate three types of young animals. The thre

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Answer:

The area function is

$A=\frac{135}{2}x-\frac{9}{2}x^2$ .

The domain and range of A is $(0,15m)$ and $(0, 253.125 m^2]$ .

Step-by-step explanation:

The given length of fencing is $90 m$ .

Let the length and width of each pen be $x$ and $y$ respectively as shown in the figure.

As there are 3 pens, so, the total area,

$A= 3 xy \;\cdots (i)$

From the figure the total length of fencing is $6x+4y$ .

Here, for a significant area for the animals, $x>0$ as well as $y>0$ as $x$ and $y$ are the sides of ben.

From the given value:

$6x+4y=90\;\cdots (ii)$

$\Rightarrow y=\frac {45}{2}-\frac{3x}{2}$

Now, from equation (i)

$A=3x\left(\frac {45}{2}-\frac{3x}{2}\right)$

$\Rightarrow A=\frac{135}{2}x-\frac{9}{2}x^2\;\cdots (iii)$

This is the required area function in the terms of variable $x$ .

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$x=15-\frac{2y}{3}$

$\Rightarrow x$ [as y>0]

So, the domain of area function is $(0,15m)$ .

For the range of area function:

As $x \rightarrow 0$ or $y\rightarrow 0$ , then $A\rightarrow 0$ [from equation (i)]

$\Rightarrow A>0$

Now, differentiate the area function with respect to $x$ .

$\frac {dA}{dx}=\frac{135}{2}-9x$

Equate $\frac {dA}{dx}$ to zero to get the extremum point.

$\frac {dA}{dx}=0$

$\Rightarrow \frac{135}{2}-9x=0$

$\Rightarrow x=\frac{15}{2}$

Check this point by double differentiation

$\frac {d^2A}{dx^2}=-9$

As, $\frac {d^2A}{dx^2}$ , so, point $x=\frac{15}{2}$ is corresponding to maxima.

Put this value back to equation (iii) to get the maximum value of area function. We have

$A=\frac{135}{2}\times \frac {15}{2}-\frac{9}{2}\times \left(\frac {15}{2}\right)^2$

$\Rightarrow A=253.125 m^2$

Hence, the range of area function is $(0, 253.125 m^2]$ .

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3 years ago