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velikii [3]
3 years ago
8

HELP PLEASE!!! WILL BE GIVING BRAINLIEST!!! (its ok if you don't get all I just need at least nine)

Mathematics
1 answer:
iogann1982 [59]3 years ago
6 0

Answers:

1.) x\leq 2

2.) y < -\frac{14}{3}

3.) x>2

4.) z\geq 9

5.) v (That variable is a "v")

6.) x=18

7.) x\geq 5

8.) y>-\frac{1}{2}

9.) x\leq 5

10.) x=12

(I answered the questions in order from top to bottom, hope I helped!)

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Solve for x in the equation.
labwork [276]

Answer:

The solution is \displaystyle x=1\pm \sqrt{47}. Fourth option

Explanation:

Solve for x:

2x^2+3x-7=x^2+5x+39

Move all the terms from the right to the left side of the equation, a zero in the right side:

2x^2+3x-7-x^2-5x-39=0

Join all like terms:

x^2-2x-46=0

The general form of the quadratic equation is:

ax^2+bx+c=0

Solve the quadratic equation by using the formula:

\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

In our equation: a=1, b=-2, c=-46

Substituting into the formula:

\displaystyle x=\frac{-(-2)\pm \sqrt{(-2)^2-4(1)(-46)}}{2(1)}

\displaystyle x=\frac{2\pm \sqrt{4+184}}{2}

\displaystyle x=\frac{2\pm \sqrt{188}}{2}

Since 188=4*47

\displaystyle x=\frac{2\pm \sqrt{4*47}}{2}

Take the square root of 4:

\displaystyle x=\frac{2\pm 2\sqrt{47}}{2}

Divide by 2:

\displaystyle x=1\pm \sqrt{47}

First option: Incorrect. The answer does not match

Second option: Incorrect. The answer does not match

Third option: Incorrect. The answer does not match

Fourth option: Correct. The answer matches exactly this option

8 0
3 years ago
What does this mean?
katrin2010 [14]

Answer:

huh that's a wierd question its giving u

3 0
3 years ago
Read 2 more answers
Find the midpoint of a segment whose endpoints are (9, –7) and (–3, 5).
suter [353]
For this question, you would have to use the midpoint formula. 
(X1 +X2 / 2 , Y1 + Y2 / 2)

In other words, 
(9 + -3 / 2 , -7 +5 / 2)
(6 / 2 , -2 / 2)
(3 , -1)

Your midpoint is (3, -1)
7 0
3 years ago
Please answer this correctly
Evgesh-ka [11]

Answer:

1/9

Step-by-step explanation:

The die has sides 1,2,3,4,5,6

P( less than 5) = 1,2,3,4 / total

                        =4/6 = 2/3

Then rolling again

P(greater than 5) = 6/total = 1/6

P( less than 5,greater than 5) = 2/3* 1/6= 1/9

3 0
3 years ago
Will mark brainliest
faust18 [17]

x^3 is strictly increasing on [0, 5], so

\max\{x^3 \mid 0\le x\le5\} = 5^3 = 125

and

\min\{x^3 \mid 0 \le x\le5\} = 0^3 = 0

so the integral is bounded between

\displaystyle \boxed{0} \le \int_0^5x^3\,dx \le \boxed{125}

8 0
2 years ago
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