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AlekseyPX
3 years ago
15

Jeanine is twice as old as her brother Marc. If the sum of their ages is 24, how old is Jeanine

Mathematics
2 answers:
Vesna [10]3 years ago
5 0
Her brother is 8: 8 x 2 is 16, and 16 + 8= 24.
docker41 [41]3 years ago
5 0
First make an equation system for the problem
An equation for "<span>Jeanine is twice as old as her brother Marc" is
</span>⇒ j = 2m
An equation for "<span>the sum of their ages is 24" is
</span>⇒ j + m = 24
<span>
Second solve the equations with subtitution method.
Subtitute j with 2m in the second equation
j + m = 24
2m + m = 24
3m = 24
m = 24/3
 m = 8
Marc is 8 years old

Find Jeanine's age by subtitute m with 8 in the first equation
j = 2m
j = 2(8)
j = 16

Jeanine is 16 years old
</span>
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3 years ago
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5. Write down the nth term of each of the following G.Ps
yawa3891 [41]

Answer:

1i) nth term = 12(-b/4)^(n-1)

ii) nth term = 3(-1/9)^(n-1)

2i) Sixth term= (-3b^5)/256

ii) Eight term = -1/1594323

Question:

The complete question as found on brainly (question-10746111):

Write down the nth term of each of the following G.Ps whose first two terms are given as follows. Also find the term stated besides each G.P.

i) 12,-3b,......sixth term

ii) 3,-1/3,..........;8th term?

Step-by-step explanation:

1) To solve terms involving Geometric progressions (GPs), we would first state the variables that are to be incorporated in the formula.

i) 12,-3b,......;sixth term

1st term = a= 12

r = common ratio = 2nd term /1st term

r = -3b/12 = -b/4

Then we would find the nth term

See attachment for details

ii) 3,-1/3,..........;8th term

1st term = a= 3

r = common ratio = 2nd term /1st term

r = (-⅓)/3 = (-⅓)(⅓) = -1/9

Then we would find the nth term

See attachment for details

2) we are to determine the sixth term in (i) and eight term in (ii)

See attachment for details

Sixth term= (-3b^5)/256

Eight term = -1/1594323

7 0
3 years ago
8x+3y=4 <br> -7x+5y=-34 <br><br> Elimination using multiplication
irga5000 [103]

Answer:

x= 2    and y = -4

Step-by-step explanation:

8x + 3y = 4 ---------------------------------(1)

-7x + 5y = -34 -----------------------------(2)

Multiply through equation (1) by 5 and multiply through equation(2) by 3

40x + 15y = 20 ----------------------------(3)

-21x + 15y =-102----------------------------(4)

Subtract equation (4) from equation (3)

61x = 122

Divide both-side of the equation by 61

61x/61 = 122/61

(At the left-hand side of the equation 61 will cancel-out 61 leaving us with just x, while at the left-hand side of the equation 122 will be divided by 61)

x = 122/61

x=2

Substitute x= 2 into equation (1)

8x + 3y = 4

8(2) + 3y = 4

16 + 3y = 4

Subtract 16 from both-side of the equation

16-16 + 3y = 4-16

3y = -12

Divide both-side of the equation by 3

3y/3 = -12/3

y = -4

x= 2    and y = -4

4 0
4 years ago
Read 2 more answers
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