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PolarNik [594]
2 years ago
11

How to solve 52+ 58+ q=180 for q.

Mathematics
2 answers:
Mama L [17]2 years ago
6 0

Answer:

70

Step-by-step explanation:

180 - 52 - 58 = q

q = 70

hope this helps...

Mice21 [21]2 years ago
3 0
<h2>Hey there! </h2><h2>\bold{52+58+q=180}\\\bold{COMBINE\ your\ LIKE\ TERMS}\\\bold{52+15=110}\\ \bold{New\ equation: 110+q=180}\\\bold{SUBTRACT\ 110\ to\ BOTH\ SIDES}\\\bold{q+110-110=180-110}\\\bold{Cancel\ out: 110-110\ because\ that\ gives\ 0}\\\bold{Keep: 180-110\ because\ it\ helps\ us\ solve\ for\ q}\\\bold{TEMPORARY \ equation: q = 180-110}\\\bold{Solve\ above\ \uparrow \ and\ you\ have\ q}\\\\\\\\\\\boxed{\boxed{\bold{Answer: q=70}}}\huge\checkmark</h2><h2 /><h2>Good luck on your assignment and enjoy your day!</h2><h2 /><h3>~\frak{LoveYourselfFirst:)}</h3>
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This is for union and intersection. I need help.
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What is the coefficient of the term of degree 8 in the polynomial below?
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What is the value of the number 4 on 264,807
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3 years ago
Show solutions please
slega [8]

Answer:

1. Their ages are:

Steve's age = 18

Anne's age = 8

2. Their ages are:

Max's age = 17

Bert's age = 11

3. Their ages are:

Sury's age = 19

Billy's age = 9

4. Their ages are:

The man's age = 30

His son's age = 10

Step-by-step explanation:

1. We make the assumption that:

S = Steve's age

A = Anne's age

In four years, we are going to have:

S + 4 = (A + 4)2 - 2 = 2A + 8 - 2

S + 4 = 2A + 6 .................. (1)

Three years ago, we had:

S - 3 = (A - 3)3

S - 3 = 3A - 9

S = 3A - 9 + 3

S = 3A – 6 …………. (2)

Substitute S from (2) into (1) and solve for A, we have:

3A – 6 + 4 = 2A + 6

3A – 2A = 6 + 6 – 4

A = 8

Substitute A = 8 into (3), we have:

S = (3 * 8) – 6 = 24 – 6

S = 18

Therefore, we have:

Steve's age = 18

Anne's age = 8.

2. We make the assumption that:

M = Max's age

B = Bert's age

Five years ago, we had:

M - 5 = (B - 5)2

M - 5 = 2B - 10 .......................... (3)

A year from now, it will be:

(M + 1) + (B + 1) = 30

M + 1 + B + 1 = 30

M + B + 2 = 30

M = 30 – 2 – B

M = 28 – B …………………… (4)

Substitute M from (4) into (3) and solve for B, we have:

28 – B – 5 = 2B – 10

28 – 5 + 10 = 2B + B

33 = 3B

B = 33 / 3

B = 11

If we substitute B = 11 into equation (4), we will have:

M = 28 – 11

M = 17

Therefore, their ages are:

Max's age = 17

Bert's age = 11.

3. We make the assumption that:

S = Sury's age

B = Billy's age

Now, we have:

S = B + 10 ................................ (5)

Next year, it will be:

S + 1 = (B + 1)2

S + 1 = 2B + 2 .......................... (6)

Substituting S from equation (5) into equation (6) and solve for B, we will have:

B + 10 + 1 = 2B + 2

10 + 1 – 2 = 2B – B

B = 9

Substituting B = 9 into equation (5), we have:

S = 9 + 10

S = 19

Therefore, their ages are:

Sury's age = 19

Billy's age = 9.

4. We make the assumption that:

M = The man's age

S = His son's age

Therefore, now, we have:

M = 3S ................................... (7)

Five years ago, we had:

M - 5 = (S - 5)5

M - 5 = 5S - 25 ................ (8)

Substituting M = 3S from (7) into (8) and solve for S, we have:

3S - 5 = 5S – 25

3S – 5S = - 25 + 5

-2S = - 20

S = -20 / -2

S = 10

Substituting S = 10 into equation (7), we have:

M = 3 * 10 = 30

Therefore, their ages are:

The man's age = 30

His son's age = 10

3 0
3 years ago
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