1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
MrRa [10]
4 years ago
13

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Mathematics
2 answers:
Strike441 [17]4 years ago
8 0

Answer:  C) x = 2 and x = -2

<u>Step-by-step explanation:</u>

Vertical Asymptotes are restrictions on the x-values.

Since the denominator cannot equal zero, then any x-value that makes the denominator equal to zero is a vertical asymptote.

x² - 4 = 0

(x - 2)(x + 2) = 0

x - 2 = 0    x + 2 = 0

  x = 2        x = -2

irakobra [83]4 years ago
6 0

Answer:

C. x = 2, and x = -2

You might be interested in
A sequence of numbers in which the ratio between any two consecutive numbers is a constant is called a(n):
Anestetic [448]
It is called a geometric sequence
6 0
3 years ago
Read 2 more answers
Alex scored 98,72, and 87 on his first three math test. What must he score on the next test to have an average of at least 86 (P
ss7ja [257]

Answer:

\huge\boxed{\text{At least an } 87\%}

Step-by-step explanation:

In order to find what score Alex must earn to have an average of 86 on all his tests, we need to first note what the formula to find the average of a data set is.

\displaystyle \frac{x_1+x_2+...x_n}{n}

What the formula means is that we have to add up all the values then divide by the total number of values.

Let's represent our unknown number as x.

\displaystyle \frac{98+72+87+x}{4} \geq 86 (since we want AT LEAST an 86).

Let's solve this inequality for x.

  • \frac{98+72+87+x}{4} \cdot 4 \geq 86 \cdot 4
  • 98+72+87+x \geq 344
  • 257 + x \geq 344
  • 257 + x  - 257 \geq 344 - 257
  • x \geq 87

So, Alex must score at least an 87% on his next quiz to have an average of 86%.

Hope this helped!

8 0
3 years ago
Doug checks census data for his city for a school project he's doing. He wants to show how many people live in the 5 areas of th
scoundrel [369]

Answer:

i have no idea sorry but there are way tooooooooooo many words in this question so it doesnt really make much sence especially the bottom part if the question.

4 0
2 years ago
Read 2 more answers
What are the names of the segments in the figure
Alborosie

Answer:

C. The three segments are segments AB, CA, and AC

7 0
3 years ago
If the cost price of 21 articles is equal to the selling price of 18 articles, find the profit percentage​
Ksenya-84 [330]

Answer:  14.29%

Step-by-step explanation:

Let C be the cost of 1 article and S the selling price for 1 article.  The profit, P, for 1 article would be:

P = S - C

For n articles, the total profit would be

nP = nS - nC

Profit percentage would be (P/S)*100%

---------------------------------------------------------

We know that 21C = 18S:  This can be written as C = (18/21)S

P = S - C

P = S - (18/21)S

P = (3/21)S or P=(1/7)S

Profit percentage would therefore be

(P/S)*100%

((1/7)S/S)*100%

(1/7)*100%

= 14.29%

5 0
3 years ago
Other questions:
  • I really need this answer to be correct
    14·1 answer
  • PLEASE ANSWERRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR
    14·1 answer
  • 14) The temperature at dawn was -5 F
    6·2 answers
  • How do you complete it
    6·1 answer
  • Need help ASAP!!!<br> WIll make you brainlist
    13·2 answers
  • Find the coordinatevof x and y of the point m wich divides the line segment m1m2 in the ratio w1:w2​
    11·1 answer
  • ASAP<br> what is the GCF between 20, 24, and 32?
    7·2 answers
  • Which fraction represents the probalility of a likely event
    8·1 answer
  • The sale price of every item in a store is 85% of it's usual price. The usual price of a backpack is 30$, what is it's sale pric
    13·1 answer
  • The average height of boys at Greg's school is 5'9" with a
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!