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lions [1.4K]
3 years ago
14

Help fast please and good night also happy Valentine's day​

Mathematics
2 answers:
Mashcka [7]3 years ago
4 0
HAPPY VALENTINES DAYYY

imma show work for each problem

1. 9c+1=10
-1 -1
----------------
9c=9
divide both sides by 9
c=1

2. 6y-5=7
+5 +5
-----------------
6y=12
divide 6 by both sides
y=2

3. 8=3a-4
+4 +4
_________
12=3a
divide 3 by both sides
4=a

4. m/5+9=11
-9 -9
--------------------
m/5=2
multiply 5 (denominator) by both sides)
m=10 cause 5s cancel out on left side

5. 13+7x=27
-13 -13
___________
7x=14
divide both sides by 7
x=2

6. 17-q=6
-17 -17
----------------
-q=-11
divide both sides by -1
q=11

7.) n-31/4=2
multiply 4 (denominator by both sides)
4's cancel out on left side
n-31=8
+31 +31
---------------
n=39


8. 1+2r=35
-1 -1
----------------
2r=34
divide 2 by both sides
r=17

9. 42+5t=8t
-5t -5t
-------------------
42=3t
divide 3 by both sides
14=t

10. 4p-3=17
+3 +3
------------------
4p=20
divide both sides by 4
p=5

hope this helps, if you have any questions just ask
-Dominant- [34]3 years ago
3 0

Answer:

#1 c=1

#2 y=2

#3 a=4

#4 m=10

#5 x=2

#6 q=11

#7 n=27

#8 r=17

#9 1st t=6 2nd t=9

#10 p=5

Did that help???

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Find two numbers whose sum is 18 who’s difference is 22
wel

Answer:

The Answer is: The first number is 20 and the second number is -2.

Step-by-step explanation:

Let f = first number and s = second number.

The sum is 18:

f + s = 18

Solve for s:

s = 18 - f

The difference is 22:

f - s = 22

Substitute:

f - (18 - f) = 22

f - 18 + f = 22

2f = 22 + 18

2f = 40

f = 20, the first number.

Solve for s:

s = 18 - f

s = 18 - 20 = -2, the second number.

Proof:

20 + (-2) = 18

20 - 2 = 18

18 = 18

Then:

f - s = 22

20 - (-2) = 22

20 + 2 = 22

22 = 22

Hope this helps! Have an Awesome Day!! :-)

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What is the sum of the first 30 terms of the arithmetic series 27,32,37,42,47
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Answer:

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Step-by-step explanation:

The formula for the sum of an arithmetic series is given by the formula:

S=\frac{k}{2}(a+x_k)

Where k is the number of terms, a is the first term, and x_k is the last term. In this case, the last term is the 30th term because we are finding the sum of the first 30 terms.

So, we need to find the 30th term. To do so, we can write an explicit formula for our sequence.

The standard form for an arithmetic sequence is given by the formula:

x_n=a+d(n-1)

Where a is the initial term and d is the common difference.

From our sequence, we can see that the initial term a is 27. The common difference is 5 because each term is 5 greater than the previous one.

So, our equation is:

x_n=27+5(n-1)

So, to find the 30th term, substitute 30 for n:

x_{30}=27+5(30-1)

Subtract:

x_{30}=27+5(29)

Multiply:

x_{30}=27+145

Add:

x_{30}=172

So, the 30th term is 172.

Now, substitute this into our original sum formula:

S=\frac{k}{2}(a+x_k)

Substitute 30 for k (the amount of terms), 27 for a (the first term), and 172 for x_k (the 30th and last term). So:

S=\frac{30}{2}(27+172)

Reduce and add:

S=15(199)

Multiply:

S=2985

So, the sum of the first 30 terms is 2985.

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