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kow [346]
3 years ago
9

Someone plz help me

Mathematics
1 answer:
sashaice [31]3 years ago
8 0

Answer:

1) n + 7 = 20

2) x + 7 = 14

3) x + n = 21

4) 11 +n = 19

5) x - 17 = 14

6) 12 - x = 8

7) x - 4 = 30

8) 8x = 56

9) 1/2n = 18

10) 2(n > 4) = 12

11) 4n = 12

12) n + 2 = 14

13) n - 10 = 4

14) n - 2 = 8

15) n - 2 = 9

16) 2n - 3 = 17

17) 2n - 10 = 22

18) 3n - 6 = 13

19) 2n - 9 = 7

20) 2(n < 3) = 13

Step-by-step explanation:

I'm not entirely sure about numbers 10 and 20..

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Please help me with this question.
Rasek [7]
The answer is 4,000 ft squared because since the sides are 100 ft and 80ft then we need to multiply them which is 8,000 then divide that by 2 since it’s a triangle which is 4,000 ft squared.
6 0
3 years ago
2(x+6)&lt;10 solve for x and inequality
Crazy boy [7]

Answer:

x<-1

Step-by-step explanation:

2(x+6)<10

Distribute 2x+12<10

Subtract 12 2x<-2

Divide 2 x<-1

4 0
3 years ago
What is the equation in slope-intercept form of the line that passes through the points (4, 19) and
mezya [45]

Answer:

y=-2/8x-3/4

Step-by-step explanation:

6 0
4 years ago
4 (w+1) =-6 gsdfnvbsfgbsdfbsdfb
Harman [31]

Answer:  Hello There!!

w = -5/2

Step-by-step explanation:

4(w+1)=−6

Step 1: Simplify both sides of the equation.

4(w+1)=−6

(4)(w)+(4)(1)=−6(Distribute)

4w+4=−6

Step 2: Subtract 4 from both sides.

4w+4−4=−6−4

4w=−10

Step 3: Divide both sides by 4.

4w/4 = -10/4

4 0
3 years ago
Read 2 more answers
Pllzzzz help asap 50 point​
Eddi Din [679]

Answer: See explanation

Step-by-step explanation:

You have to remeber that x to the power of r/q is the same thing as the qth root of x to the power of r.

For the second one, you would have the cubic root of 2 to the power of 2, or 4. So, your answer would be \sqrt[3]{4}.

For the third one, you would have the square root of 3 to the power of 3, or 27. So, your answer would be \sqrt[2]{27}.

For the fourth one, you would have the cubic root of 3 to the power of 1, or 3. So, your answer would be \sqrt[3]{3}.

Hope this helps! :)

3 0
2 years ago
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