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3 years ago

Solve for x: −2x + 5 < 7

1 answer:

7 0

-2x + 5 < 7

-2x + 5 < 7
 -5 -5 deduct 5 from both sides
-2x < 2
÷ -2 ÷ -2 divide both sides by negative 2. Because of the division
x > -1 using a negative number, the sign is then reversed.
from < it becomes >.

The value of x should be greater than -1. It can be 0, 1, 2, so on...

To check: Revert back to the original sign which is <.

x = 1
-2x + 5 < 7
-2(1) + 5 < 7
-2 + 5 < 7
3 < 7

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The students at a High School earned money for an international animal rescue foundation. 82 seniors earned an average $26.75 pe

Anon25 [30]

Answer: A. $16.13

Step-by-step explanation:

Total students = 82+74+96+99 =351

Sum of earnings of 82 seniors = $26.75 x 82= $2193.5

Sum of earnings of 74 juniors = $12.25 x 74 = $906.5

Sum of earnings of 96 sophomores = $15.50 x 96 = $1488

Sum of earnings of 99 freshmen = $10.85 x 99 = $1074.15

Total earnings = $2193.5 + $906.5+ $1488 +$1074.15

= $5662.15

(Total earnings) ÷ (Total students )

= $5662.15÷ 351

= $16.13

4 0

3 years ago

Find the first and second derivatives of the function y = sin(x^2)

exis [7]

Y = sin(x^2)

Use the chain rule.

We want dy/dx.

dy/dx = cos(x^2)*2x

dy/dx = 2xcos(x^2)

Did you follow?

7 0

2 years ago

Pls help me with this proof! I only have a short amount of time

Anna35 [415]

Because it is in the middle of those given points

7 0

2 years ago

Simplify -36 + x < 8

krok68 [10]

Step-by-step explanation:

add 36 to both sides to cancell out -36

-36+36=0

8+36=44

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so all values for x are less than 44

Hope that helps :)

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3 0

2 years ago

The county hospital is located at the center of a square whose sides are 3 miles wide. If an accident occurs within this square,

Dmitry_Shevchenko [17]

Answer:

Step-by-step explanation:

The x-coordinate of the accident is a random variable of uniform distribution with range [-1.5,1.5]. The same can be said for the y-coordinate. However, |x| and |y| are also uniform variables with range [0,1.5], since we still have the same probability on every pair of intervals with the same lenght.

Note that |X| and |Y| are independent with each other, therefore E(|x|+|y|) = E(|x|) + E(|y|) = 2*E(|x|). The last equality holds because both x and y are identically distributed, then so are |x| and |y|.

E(|x|) = $\int\limits^{1.5}_0 {\frac{x}{1.5} \, dx = \frac{2}{3}* ((1.5^2)/2 - 0^2/2) = \frac{3}{2}$

Therefore, E(|x| + |y|) = 2 * 3/2 = 3.

I hope that works for you!

4 0

3 years ago