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svetlana [45]
3 years ago
8

5 years ago, the age of a man was 7 times the age of his son. After five years, the age of the man will be 3 times the age of hi

s son from now. How old are the man and the son now?
Mathematics
1 answer:
Feliz [49]3 years ago
3 0

Answer:

10 years

40 years

Step-by-step explanation:

let present ag e of son=x

5 years ago age of son=x-5

5 years ago age of man=7(x-5)=7x-35

present age of man=7x-35+5=7x-30

after 5 years

age of son=x+5

age of man=7x-30+5=7x-25

also 7x-25=3(x+5)

7x-25=3x+15

7x-3x=15+25

4x=40

x=10

age of son=10 years

age of man=7*10-30=70-30=40 years

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Salsk061 [2.6K]

Answer:

B

Step-by-step explanation:

Area of pool = l × b

l = 6x and b = 3x - 5

Area = 6x(3x - 5)

Area = 18x² - 30x

4 0
3 years ago
1.75x+1.25y=2<br> 7x+5y=9<br> how many solutions does this system of equations have?
Vlad1618 [11]

Answer:

true 7x+5y=9

Step-by-step explanation:

7 0
3 years ago
What are the x- and y-intercepts of the line?. 2x – 3y = 12. A. x-intercept –6, y-intercept 4. B. x-intercept –4, y-intercept 6.
tigry1 [53]
D. x-intercept 6, y-intercept –4.
7 0
3 years ago
Read 2 more answers
If a=under root s(s-a)(s-b)(s-c) then the value of a, when a =3 b =4c = 5 and s =a+b+c by 2
kramer

Answer:

6

Step-by-step explanation:

Here it's given that ,

\sf\red{\longrightarrow} A=\sqrt{s(s-a)(s-b)(s-c)}

Also ,

\sf\red{\longrightarrow} s =\dfrac{a+b+c}{2}

And we need to find out the value of A , when

  • a = 3
  • b = 4
  • c = 5

So , on substituting the respective values to find s we have ,

\sf\red{\longrightarrow} s =\dfrac{3+4+5}{2}=\dfrac{12}{2}=\bf{6}

Now let's find out A as ,

\sf\red{\longrightarrow} A = \sqrt{6(6-3)(6-4)(6-5)} \\

\sf\red{\longrightarrow}A =\sqrt{6 * 3 * 2 * 1}\\

\sf\red{\longrightarrow} A = \sqrt{3^2 * 2^2 *1^2}\\

\sf\red{\longrightarrow}A = 3*2\\

\sf\red{\longrightarrow}\boxed{\qquad \blue{\bf A = 6 \qquad}}

7 0
2 years ago
Find the next 3 terms in the geometric sequence -36, 6, -1, 1/6, ...
timama [110]

we are given

geometric sequence -36, 6, -1, 1/6, ...

first term is -36

a_1=-36

now, we can find common ratio

r=\frac{6}{-36}

r=-\frac{1}{6}

now, we can find nth term

a_n=a_1(r)^{n-1}

now, we can plug values

and we get

a_n=36(-\frac{1}{6})^{n-1}

now, we can find 5th term , 6th term, 7th term

fifth term:

a_5=36(-\frac{1}{6})^{4}

a_5=\frac{1}{36}

sixth term:

a_6=36(-\frac{1}{6})^{5}

a_6=-\frac{1}{216}

seventh term:

a_7=36(-\frac{1}{6})^{6}

a_7=\frac{1}{1296}

so, next terms are

a_5=\frac{1}{36} , a_6=-\frac{1}{216}

, a_7=\frac{1}{1296}.............Answer

6 0
3 years ago
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