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3 years ago

Write an equation in slope-intercept form

Mathematics

1 answer:

PolarNik [594]3 years ago

6 0

OMG im suffering in this too

but im starting to get this so lets take a look

i tried writing an equation but all i get is one so

answer is one and slope is 0 hope i could help

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A(n) = -5 + 6(n - 1)

Andreas93 [3]

Answer:

domain is all reall numbers

3 0

3 years ago

Read 2 more answers

Hamburger Hut sells regular hamburgers as well as a larger burger. Either type can include cheese, relish, lettuce, tomato, must

Studentka2010 [4]

Answer:

a) 40 different hamburgers can be ordered with exactly three extras

b) 20 different regular hamburgers can be ordered with exactly three extras

c) 7 different regular hamburgers can be ordered with at least five extras

Step-by-step explanation:

The order in which the extras are ordered is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem:

2 options of hamburger(regular or larger)

6 options of extras(cheese, relish, lettuce, tomato, mustard, or catsup.).

(a) How many different hamburgers can be ordered with exactly three extras?

1 hamburger type, from a set of 2.

3 extras, from a set of 6. So

$C_{2,1}*C_{6,3} = \frac{2!}{1!(2-1)!}*\frac{6!}{3!(6-3)!} = 2*20 = 40$

40 different hamburgers can be ordered with exactly three extras

(b) How many different regular hamburgers can be ordered with exactly three extras?

3 extras, from a set of 6. So

$C_{6,3} = \frac{6!}{3!(6-3)!} = 20$

20 different regular hamburgers can be ordered with exactly three extras

(c) How many different regular hamburgers can be ordered with at least five extras?

Five extras:

5 extras, from a set of 6. So

$C_{6,5} = \frac{6!}{5!(6-5)!} = 6$

Six extras:

6 extras, from a set of 6. So

$C_{6,6} = \frac{6!}{6!(6-6)!} = 1$

6 + 1 = 7

7 different regular hamburgers can be ordered with at least five extras

8 0

3 years ago

As the limit of x goes to 3 from the left find the limit of (x/ sqrt x^2 -9)

Dimas [21]

$\displaystyle\lim_{x\to3^-}\frac x{\sqrt{x^2-9}}=\lim_{x\to3^-}\frac x{\sqrt{x^2}\sqrt{1-\frac9{x^2}}}=\lim_{x\to3^-}\frac x{|x|\sqrt{1-\frac9{x^2}}}$

Since $x>0$ , you have $|x|=x$ and the limand reduces to

$\displaystyle\lim_{x\to3^-}\frac1{\sqrt{1-\frac9{x^2}}}$

For $x>3$ , you have $\sqrt{1-\frac9{x^2}}>0$ since $\dfrac9{x^2}$ will always be smaller than 1, which means $\dfrac1{\sqrt{1-\frac9{x^2}}}\to+\infty$