Ask question

Ask question

All categories

SCORPION-xisa [38]

3 years ago

10

A store is having a going -out-of buisness sale.a television set that orginally cost 3000 has been marked down by30% what do you

pay for the television set with the markdown and 7% tax

1 answer:

allsm [11]3 years ago

7 0

2,247

Step-by-step explanation:

take 3000 and minus that by 30 percent, then after that, add 7 percent to get your total.

You might be interested in

Which of the following tables does NOT define a function?

madreJ [45]

Answer:

c

Step-by-step explanation:

doesnt define a function

5 0

3 years ago

Read 2 more answers

School x and school y play each other in a competition school x has 8 more points then school y school x has three times as many

Nookie1986 [14]

I think the anwser is team y has 4 points and team x has 12 points because 4×3=12-4=8 so team x has 3× the points team y does and has an 8 point lead.
hope this helps and dont forget to hit that heart :)

5 0

3 years ago

Read 2 more answers

What is the constant of proportionality between y and x the graph?

olga55 [171]

I believe it’s the last option 1/2

4 0

3 years ago

Which choice is the equation of a line that passes through point (7,3) and is parallel to the line represented by this equation?

Julli [10]

Two lines are parallel, means, they have the same slope.
the slop is 2/7 in this case
use point-slop form to write your equation: the given point is (7,3)
(y-3)=(2/7)(x-3)
Change it to slope-intercept form if needed: y=(2/7)x+(15/7)

7 0

3 years ago

Use the properties of limits to help decide whether each limit exists. If limit exits, find its value.

Sergio039 [100]

Answer:

a) The limit doesnt exist.

b) The limit exists and its value is 1/10

Step-by-step explanation:

a) We take the highest power (x⁴) as common factor in both the numerator and the denominator.

$\lim_{x \to \infty} \frac{x^4-x^3-3x}{7x^2+9} = \lim_{x \to \infty} \frac{x^4(1-1/x-3/x^3)}{x^4(7/x^2+9/x^4)} = \lim_{x \to \infty} \frac{1-1/x-3/x^3}{7/x^2+9/x^4}$

The limit of the numerator is 1 (when x goes to infinity) and the limit on the second part is 0. Hence the limit doesnt exist (it goes to infinity).

b) Note that if we multiply √x - 5 by its conjugate of , which is √x + 5, we obtain x - 25, thus

$\lim_{x \to 25} \frac{\sqrt x - 5}{x - 25} = \lim_{x \to 25} \frac{(\sqrt{x} - 5)}{(\sqrt{x} - 5) * (\sqrt{x}+5)} = \lim_{x \to 25} \frac{1}{\sqrt x + 5} = \frac{1}{5+5} = \frac{1}{10}$

Hence, the limit exists and its value is 1/10.

6 0

3 years ago