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Mandarinka [93]
3 years ago
8

Which of the x-values are solutions to the following inequality? x< 100 Help ASAP

Mathematics
1 answer:
netineya [11]3 years ago
3 0

Answer:

x = - infinity to 99

Step-by-step explanation:

The values of x must less than 100.

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Solve system of equation using elimination by addition.<br><br> Will give brainliest!!
DedPeter [7]

Answer:

x = 2, y = -3

Step-by-step explanation:

Add the 2 equations together, then solve for x.

(2x + 2y) + (3x - 2y) = (-2) + (12)\\ 2x + 3x + 2y - 2y = -2 + 12\\5x = 10\\x = 2

Now that we know the value of x, we can easily find y by substitution.

3x - 2y = 12\\3(2) - 2y = 12\\6 - 2y = 12\\-2y = 6\\y = -3

These are the answers

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2 years ago
G ^-1(3)=<br><br><br> A.-1<br> B.0<br> C.3<br> D. 5<br><br> Please help me understand this!!
Natasha2012 [34]
The answer the 3,

First you have to apply the exponent rule which is 1 - 3• 1/g

Then you multiply the fractions - 1•3/g

Then multiply the numbers 1 and 3, which is 3, so there you have it 3/g, or 3, they’re the same thing.
8 0
3 years ago
Percentage grade averages were taken across all disciplines at a particular university, and the mean average was found to be 83.
Paha777 [63]
Correct Answer:
Option A. 0.01

Solution:
This is a problem of statistics and uses the concept of normal distributions. We need to convert the score of 90 into z-score and then find the desired probability from standard normal distribution table.

Converting 90 to z-score:

z= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10}} }=2.33

Now we are to find the probability of z score being more than 2.33. From the z-table the probability comes out to be 0.01.

Therefore, we can conclude that the probability of class average is greater than 90 is 0.01.
4 0
2 years ago
Malia and her sister watched an action movie that was 2 hours and 5 minutes long. After
Black_prince [1.1K]

Answer: they started at 1:10 pm

7 0
3 years ago
Read 2 more answers
Find the value of x and simplify completely.
jok3333 [9.3K]

Answer:

<h2>x=9√10</h2>

Given: A right triangle in which an altitude is drawn from the right angle vertex to the hypotenuse.

To find: 'x' the larger leg of triangle

Solution,

Using let rule for similarity in right triangle:

\frac{leg}{part}  =  \frac{hypotenuse}{leg}  \\ or \:  \frac{x}{27}  =  \frac{3 + 27}{x}  \\ or \: x \times x = 27(3 + 27) \\ or \: x \times x = 81 + 729 \\ or \:  {x}^{2}  = 810 \\ or \:  {x}^{2}  = 81 \times 10 \\ or \:  {x}  =  \sqrt{81 \times 10}  \\ or \: x =  \sqrt{81}  \times  \sqrt{10}  \\ or \: x =  \sqrt{ {(9)}^{2} }  \times  \sqrt{10}  \\  \: x = 9 \sqrt{10}

Hope this helps...

Good luck on your assignment..

7 0
3 years ago
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