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gogolik [260]
3 years ago
9

List the intergers that satisfy both these inequalities 2x+9<0 and x>-12

Mathematics
2 answers:
pogonyaev3 years ago
7 0

Answer:

The integers that satisfy both inequalities are -5, -6, -7, -8, -9, -10 and -11.

Step-by-step explanation:

Firstly, you have to solve the inequality :

2x + 9 < 0

2x <  - 9

x <  -  \frac{9}{2}

x <  - 4.5

Next, given that x is greater than -12 but smaller than -4.5 so the inequality for x is :

- 12 < x <  - 4.5

Integer is a <u>positive</u> or a <u>negative</u> number but <em>'</em><em>cannot be written in the form of decimal</em><em>'</em>. So the integers that satisfy both inequalities are :

- 12 < x <  - 4.5

x =  - 11, - 10, - 9, - 8, - 7, - 6, - 5

hammer [34]3 years ago
3 0

Answer:

-11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1

(would really, reallly appreciate the brainliest)

Step-by-step explanation:

2x+9 < 9

can be rewritten as

2x < 0

and therefore

x < 0

the other inequality states that x hat to be bigger then -12

so all integers that satisfy -12 < x < 0 are in the set we search, but be aware that -12 and 0 don't fall into this set, bc -12 for example isn't bigger than -12

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What is the slope? I'll marl the first correct answer brainliest. Please.
Leno4ka [110]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Will someone plz help me this us due and I am confused
mihalych1998 [28]

Hey there! I would love to help you.

Question 1: In both questions, we have to find the repeating decimal as a fraction. There is a specific way to find the fraction given a repeated decimal. In our equation, our variable x will represent the fraction.

x=0.272727....

If you know how too solve systems of equations by elimination, we need to to do something similar to eliminate all of the repeating parts so we can solve for x.

We need to make this a number greater than zero but line up the digits so that we can eliminate every single repeating digit.

To do this, we move the digit as many times to the right as there are digits that repeat. In this case, there are two repeating digits, so we multiply the whole equation by 100.

100x=27.2727...

Now, we can subtract the first equation from the second  so that the repeating parts are removed and we can then solve for x and find our fraction.

99x=27

Now, we solve for x.

x=27/99

Now we subtract this from the first fraction.

11/6-27/99= 1 37/66

To make this into a repeating decimal, we just divide the numerator by the denominator.

1.560606060...

As we can see, the 60 is repeating, so the answer is the second option on the first one.

Question 2: In this case, we have two repeating decimals. Let's solve for them both. Because we have a digit after the decimal that does not repeat, we need to move it before we subtract.

10x=4.090909...

We need to line up the decimals, and in this case we need to multiply by 1000.

990x=409.090909...

Now we subtract...

990x=405

We solve...

x=9/22

Now we do the other one.

10x=6.81818181...

We multiply by 100 to line up the decimals for subtraction.

1000x=681.8181....

We subtract...

990x=675

x=15/22

Now, we add our fractions.

24/22 or 1 2/22

Now we turn it back into a repeating.

24/22= 1.09090909...

For this question, select the second option.

I hope this helps!

7 0
3 years ago
What is the square root of 5?
shutvik [7]
5 x 5 = 25

answer: 25.
7 0
4 years ago
Read 2 more answers
Two real solutions of tx^2+3x_7=0
Olegator [25]

The number of solutions of a quadratic equation

ax^2+bx+c=0

Depends on its discriminant

/Delta=b^2-4ac

If /Delta>0 there are two distinct solutions

If /Delta=0 there are two coincident solutions

If /Delta<0 there are no solutions.

We know that there are two real solutions (I assume you mean distinct solutions), so we know that the discriminant is positive:

The number of solutions of a quadratic equation

Depends on its discriminant

If there are two distinct solutions

If there are two coincident solutions

If there are no solutions.

We know that there are two real solutions (I assume you mean distinct solutions), so we know that the discriminant is positive:

b^2-4ac=9+28t>0\iff t>-\dfrac[9][28]

3 0
3 years ago
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LekaFEV [45]
A 45° sector is 1/8 of the area of the circle, so is (72π m^2)/8 = 9π m^2.
4 0
3 years ago
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