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3 years ago

What is the value of x in the quadratic equations?

Mathematics

1 answer:

denis-greek [22]3 years ago

3 0

Since they have to equal each other you can do reverse math, and if I’m not wrong i think its C. x=3

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A cone has a radius of 5 ft and a height of 15 ft. It is empty and is being filled with water at a constant rate of 15 15 ft 3 f

kvasek [131]

Answer:

$\frac{dr}{dt}=0.11062ft/sec$

Step-by-step explanation:

From the question we are told that:

Radius $r=5ft$

Height $H=15ft$

Rate $R=15ft/3sec =5ft/s$

Surface Radius $R_{surf}=2.2ft$

Generally the equation for Volume is mathematically given by

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

Since radius to height ratio gives

$\frac{r}{h}=\frac{5}{15}$

$\frac{r}{h}=\frac{1}{3}$

$h=3r$

Therefore

$V=\frac{1}{3}\pi*r^2(3r)$

$V=\pi r^3$

Generally the equation for Change of Volume is mathematically given by

$\frac{dv}{dt}=\pi \frac{d}{dt}(r^3)$

$\frac{dv}{dt}=\pi 3*r^2 \frac{dr}{dt}$

$\frac{dv}{dt}=\pi 3*(2.2)^2 \frac{dr}{dt}$

$\frac{dr}{dt}=\frac{5}{45.62}$

$\frac{dr}{dt}=0.11062ft/sec$

3 0

3 years ago

Antonio has read 147 pages of a book.He has completed 70% of the book. How many more pages does he need to read to finish the bo

aleksklad [387]

210-147 so he has 63 more pages to read I just kept putting numbers in and multiplying by .7 till I got 147 from 210

6 0

4 years ago

What’s the answer please help

noname [10]

Answer:

$\boxed{\text{A. }\math{\left \{ x \, | \, x \in \mathbb{R}, x < -2 \right \}}}}$

Step-by-step explanation:

The open circle means that the point is not included in the solution set, and the arrow pointing left means that all numbers less than -2 are members.

In set-builder notation, each term has a special meaning. The braces enclose the members of the set.

Here's how you translate the notation,

$\begin{array}{rcl}\\\left \{ & = & \text{The set of}\\x & = & \text{all x values}\\| & = &\text{such that}\\x & = & x\\\in & = &\text{is a member of}\\\mathbb{R}, & = &\text{all real numbers, and}\\x < -2 & = & \text{x is less than -2}\\\end{array}\\\text{The answer is }\boxed{\textbf{A. }\mathbf{\left \{ x \, | \, x \in \mathbb{R}, x < -2 \right \} }}}$

3 0

3 years ago

This graph shows a system of equations what is the solution to the system of equations 1.(-2,2) 2.(2,2) 3.(2,-2) 4.(-2,-2)

Nutka1998 [239]

Answer:

I honestly don't know that fashoo

5 0

3 years ago

Need help in number 12 and 13 PLEASEE!! I don’t get it and I start school the day after tomorrow

Natasha_Volkova [10]

Answer:

The explanations for the graphs are provided down below. Please let me know if you have any questions about my answer.

12 and 13 as written on the worksheet is right.

Step-by-step explanation:

12) The answer given is correct.

The relation between x and y is given as:

$y=\frac{x^2}{2}-3$ with $x \in \{-4,-2,0,2}$ .

I replaced the word domain with x since the domain is the set of x's for which the relation exists.

We are going to replace x with each of the x's given to see what y corresponds to each.

Let's begin with x=-4.

$y=\frac{x^2}{2}-3$ with $x=-4$ :

$y=\frac{(-4)^2}{2}-3$

$y=\frac{16}{2}-3$

$y=8-3$

$y=5$ .

So (-4,5) is an ordered pair that should be on our graph.

To find this point you move left 4 from origin then up 5. Now you put a dot where you have landed. Your graph does show this point.

Moving on.

Let's do the next x: x=-2.

$y=\frac{x^2}{2}-3$ with $x=-2$ :

$y=\frac{(-2)^2}{2}-3$

$y=\frac{4}{2}-3$

$y=2-3$

$y=-1$ .

So (-2,-1) is an ordered pair that should be on our graph.

To find this point you move left 2 from origin and then down 1. Now you put a dot where you have landed. Your graph shows this point as well.

Now x=0.

$y=\frac{x^2}{2}-3$ with $x=0$ :

$y=\frac{0^2}{2}-3$

$y=\frac{0}{2}-3$

$y=0-3$

$y=-3$

So (0,-3) is an ordered pair that should be on our graph.

To find this point you move left and right none and down 3. Now you put a dot where you have landed. Your graph shows this point.

Now the last point will be at x=2.

$y=\frac{x^2}{2}-3$ with $x=2$

$y=\frac{2^2}{2}-3$

$y=\frac{4}{2}-3$

$y=2-3$

$y=-1$ .

So (2,-1) is an ordered pair that should be on our graph.

To find this point you move 2 units right from the origin and then down 1 unit. Now put a dot where you landed. The graph shows this point as well.

13) The answer given is correct.

$g(x)=|x|$ is the parent function and makes like a V shaped graph where it's vertex is at (0,0).

If we want to move this graph right 3 it becomes:

$m(x)=|x-3| \text{ or } m(x)=|(-1)(-x+3)|=|-1||-x+3|=1|-x+3|=|-x+3|=|3-x|$ .

If you move that up once it becomes:

$n(x)=|x-3|+1$ or $n(x)=|3-x|+1$ which is the curve given.

If you don't know about transformations you can choose a few points to plug in to see what's going on with the graph.

Let's choose x=-5,-3,-1,0,1,3,5.

x=-5

f(-5)=|3--5|+1=|3+5|+1=|8|+1=8+1=9.

There is no room for (-5,9) on our graph but if you extended the left hand side of the absolute value function there you would see that (-5,9) is reached.

x=-3

f(-3)=|3--3|+1=|3+3|+1=|6|+1=6+1=7.

(-3,7) should be a point on the graph. Same thing for this point as (-5,9).

x=-1

f(-1)=|3--1|+1=|3+1|+1=|4|+1=4+1=5.

(-1,5) is located on the graph.

x=0

f(0)=|3-0|+1=|3|+1=3+1=4.

(0,4) is also located on the graph.

x=1

f(1)=|3-1|+1=|2|+1=2+1=3.

(1,3) is located on the graph.

x=3

f(3)=|3-3|+1=|0|+1=0+1=1.

(3,1) is located on the graph.

x=5

f(5)=|3-5|+1=|-2|+1=2+1=3.

(5,3) is located on the graph.

Now if we weren't given the graph already:

I would plot the points I found which were:

(-5,9)

(-3,7)

(-1,5)

(0,4)

(1,3)

(3,1)

(5,3)

We should get a basic idea of what the function looks like from these points.

I will graph them. You will have to connect these points because the domain isn't discrete like number 12 is. That is they didn't list out elements for the domain.

I'm going to graph one more point after x=5.

How about x=7?

$f(7)=|3-7|+1=|-4|+1=4+1=5$

So (7,5) is also a point on the graph.

You should see that the blue points are following the red path I made there.

8 0

3 years ago