Question 18: Please help, I'm not sure but I think the answer is either C or D.

The cost of a movie ticket is increased by 15%. The old price was five dollars how much are they now?

Grace [21]

Answer:

5.75 is the answer.

$5.00 x 0.15=0.75

$5.00+0.75=$5.75

4 0

3 years ago

Write the equation of a line that includes the point (-4,0) and has a slope of -3/5 in slope intercept form

dolphi86 [110]

In "slope-intercept form"
y = mx +b
the value "m" is called the slope, and the value "b" is called the intercept.

There is another form for the equation of a line, called "point-slope form".
y = m(x -h) +k
where m is still the slope and (h, k) correspond to the (x, y) of the point.

If you write the equation of your line in this "point-slope form", it is easily manipulated to be in the "slope-intercept form".
Fill in
m = (-3/5)
h = -4
k = 0
y = (-3/5)(x -(-4)) +0

Now, you simplify this by using the distributive property.
y = (-3/5)x -(3/5)*4
y = (-3/5)x -12/5 . . . . . . . . . the desired equation

_____
Your understanding of math improves immensely when you become familiar with the terminology. A lot of the rest of it is pattern matching--identifying the parts of one expression that correspond to the parts of another one.

(You will see another version of the "point-slope form", but I find this one the easiest to use for manipulating the equation to other forms.)

5 0

4 years ago

Kiaraly answered 27 out of 30 questions correctly on her math test. Later she answered 22 of 25 questions correctly on her scien

STatiana [176]

It’s same because 30-27=3
25-22=3

5 0

4 years ago

I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.

$x\in(-\infty,0) \implies x = -1 \implies f''(-1) = 32e^2 > 0$

$x\in(0,1) \implies x = \dfrac12 \implies f''\left(\dfrac12\right) = \dfrac5{43} > 0$

$x\in(1,3) \implies x = 2 \implies f''(2) = -\dfrac{16}{e^4} < 0$

$x\in(3,\infty) \implies x = 4 \implies f''(4) = \dfrac{192}{e^8} > 0$

The sign of $f''$ changes to either side of $x=1$ and $x=3$ , but not $x=0$ . This means only $\boxed{x=1}$ and $\boxed{x=3}$ are inflection points.

4 0

1 year ago

Read 2 more answers

Ed planned to read 25 pages a day of his favorite book by some deadline. However, he read 8 pages more a day and there were just

GaryK [48]

Answer:

175 pages

Step-by-step explanation:

Let's call 'd' the number of days before the deadline, and 'n' the total number of pages of the book.

We have the following equations:

(1) $25 d = n$ --> if we multiply the number of pages per day (25) times the number of days (d), we get the total number of pages of the book (n)

(2) $(25+8)(d-2)=n-10$ --> he reads 8 pages more per day (so, 25+8), for a number of days equal to (d-2) (d is the deadline), and he reads a total of n-10 pages (because only 10 are left)

We can substitute (1) into (2) and we find:

$33(d-2)=25d-10\\33d-66=25d-10\\33d-25d=66-10\\8d=56\\d=7$

So, the deadline is 7 days, and from eq.(1) we find the number of pages in the book:

$n=25d=25 (7)=175$

8 0

3 years ago