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
Savatey [412]
3 years ago
9

Solve – 7 = √ 2x – 9.

Mathematics
2 answers:
motikmotik3 years ago
5 0
Add 9 to both sides
9-7=√2x
2=√2x
squqre both sides
4=2x
divide both sides by 2
2=x
Akimi4 [234]3 years ago
4 0

-7 =√2x -9

√2x=9-7

√2x = 2

x= 2/√2

x=√2

You might be interested in
For what values of x is the function f(x)=x^2-4x-5 increasing?
stepladder [879]
For x ∈ [ - b / ( 2a ) ; +oo ) , f is increasing; => x ∈ [ +2 ; +oo ).
6 0
3 years ago
Performance Matters
spayn [35]

Answer:

the 1st, 2nd and 6th statements are true.

and probably the 3rd statement is true.

I am not sure about the 3rd statement, as I cannot read the original in your screenshot, and your transcribed description is probably not correct and contains typos.

but all you need is in the explanation below to decide, if the actual 3rd statement is true or not.

if "58 + 79 = 1268" actually means "5s + 7g = 1268" then it is true. otherwise it is false.

Step-by-step explanation:

g = number of general tickets sold

s = number of student tickets sold

so, in total

g + s = 234

tickets were sold.

and the revenue was

7g + 5s = $1,268

out of these 2 basic equating we get

g = 234 - s

and then

7(234 - s) + 5s = 1,268

1,638 - 7s + 5s = 1,268

-2s = -370

s = 185

g = 234 - s = 234 - 185 = 49

so, we know

185 student tickets were sold for 5×185 = $925

49 general tickets were sold for 7×49 = $343

7 0
2 years ago
As the sample size becomes larger, the sampling distribution of the sample mean approaches a _____. a. binomial distribution b.
kkurt [141]

Answer:

c. normal probability distribution

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

By the Central Limit Theorem, it is a normal distribution, so option c.

7 0
2 years ago
There are 15 boys and 16 girls in Gabriel’s class. One person is chosen at random. What is the probability that the person chose
aalyn [17]

Answer:

16/31

Step-by-step explanation:

7 0
3 years ago
Help I’ll give 50 points<br> 12) 4x- y=8<br> 5x+y=1
Luba_88 [7]

12) 4x- y=8

5x+y=1

Correct.

4 0
3 years ago
Other questions:
  • Factor a number, variable, or expression out of the polynomial shown below. 15x^3 + 20x
    5·1 answer
  • If a quadrilaterals diagonals bisect its vertices, what kind of quadrilateral must it be?
    14·1 answer
  • I really need help with this since I have a math test tomorrow, I'd really appreciate it if someone could help.
    9·1 answer
  • Ken’s fish tank cost $35.30. The air pump cost $12.50. The rocks for the floor of the tank cost $3.99. Fish food cost $4.25 How
    6·1 answer
  • Parallelogram<br> 15 in<br> 18 in<br> Area:
    13·1 answer
  • Division number story with the answer 5/2
    12·1 answer
  • Which graph represents the solution to this system of inequalities:
    12·1 answer
  • Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis: y=x6, y=1 abo
    11·1 answer
  • Please hurry I need this soon and no links
    8·1 answer
  • Which values are solutions to the equation below?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!