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2 years ago

The product of a number and negative 8

Mathematics

1 answer:

notsponge [240]2 years ago

7 0

Answer:

-8x

Step-by-step explanation:

Because you don't know what that other number is, just substitue in x.

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For what values of a and m does f(x) have a horizontal asymptote at y = 2 and a vertical asymptote at x = 1? f(x)=2x^m/x+a {a =

iVinArrow [24]

Answer:

See below

Step-by-step explanation:

$f(x)=\frac{2x^{m} }{x+a}$

$f(x)=\frac{2x^{1} }{x-1}$

If x=1, then 1-1=0, which implies a vertical asymptote at x=1 since dividing by 0 is undefined.

3 0

2 years ago

Consider the square root function f(x) = sqrt x + 2 + 1 to complete the table of values below. find each missing value and then

Scrat [10]

Given the function :

$f(x)=\sqrt[]{x+2}+1$

We need to find each missing value

Given x = -3 , -2 , -1 , 2 , 7

So, substitute with each value of x to find the corresponding value of f(x)

$x=-3\rightarrow f(x)=\sqrt[]{-3+2}+1=\sqrt[]{-1}+1$

So, there is no value for f(x) at x = -3 (the function undefined because the square root of -1)

$\begin{gathered} x=-2\rightarrow f(x)=\sqrt[]{-2+2}+1=\sqrt[]{0}+1=0+1=1 \\ \\ x=-1\rightarrow f(x)=\sqrt[]{-1+2}+1=\sqrt[]{1}+1=1+1=2 \\ \\ x=2\rightarrow f(x)=\sqrt[]{2+2}+1=\sqrt[]{4}+1=2+1=3 \\ \\ x=7\rightarrow f(x)=\sqrt[]{7+2}+1=\sqrt[]{9}+1=3+1=4 \end{gathered}$

the graph of the function and the points will be as shown in the following image :

5 0

8 months ago

What is the result when 3x3 + 4x2 + 11x + 10 is divided by x + 1?

algol13

The result you get when you divide is 11 + 16/x + 1

3 0

2 years ago

Read 2 more answers

To rent a building for a school dance, Ava paid $120 plus $2.50 for each student who

Shtirlitz [24]

(325-120)/2.5=2050/25=82
=>82 students attended the dance

3 0

3 years ago

Find an asymptote of this conic section. 9x^2-36x-4y^2+24y-36=0

ki77a [65]

We will begin by grouping the x terms together and the y terms together so we can complete the square and see what we're looking at. $(9x^2-36x)-(4y^2+24y)-36=0$ . Now we need to move that 36 over by adding to isolate the x and y terms. $(9x^2-36x)-(4y^2+24y)=36$ . Now we need to complete the square on the x terms and the y terms. Can't do that, though, til the leading coefficients on the squared terms are 1's. Right now they are 9 and 4. Factor them out: $9(x^2-4x)-4(y^2-6y)=36$ . Now let's complete the square on the x's. Our linear term is 4. Half of 4 is 2, and 2 squared is 4, so add it into the parenthesis. BUT don't forget about the 9 hanging around out front there that refuses to be forgotten. It is a multiplier. So we are really adding in is 9*4 which is 36. Half the linear term on the y's is 3. 3 squared is 9, but again, what we are really adding in is -4*9 which is -36. Putting that altogether looks like this thus far: $9(x^2-4x+4)-4(y^2-6y+9)=36+36-36$ . The right side simplifies of course to just 36. Since we have a minus sign between those x and y terms, this is a hyperbola. The hyperbola has to be set to equal 1. So we divide by 36. At the same time we will form the perfect square binomials we created for this very purpose on the left: $\frac{(x-2)^2}{4}- \frac{(y-3)^2}{9}=1$ . Since the 9 is the bigger of the 2 values there, and it is under the y terms, our hyperbola has a horizontal transverse axis. a^2=4 so a=2; b^2=9 so b=3. Our asymptotes have the formula for the slope of $m=+/- \frac{b}{a}$ which for us is a slope of negative and positive 3/2. Using the slope and the fact that we now know the center of the hyperbola to be (2, 3), we can solve for b and rewrite the equations of the asymptotes. $3= \frac{3}{2}(2)+b$ give us a b of 0 so that equation is y = 3/2x. For the negative slope, we have $3=- \frac{3}{2}(2)+b$ which gives us a b value of 6. That equation then is y = -3/2x + 6. And there you go!

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2 years ago