Dad is 7 times as old as his son. 10 years later, he will be three times as old as his son.find their ages?

Answers

Answer 1

Answer: so dad is represented by x
son is represented by y

so x is (x=) 7 times his son (7y)

our first sentence is x=7y

10 years later (x+10, y+10) he will be 3 times as old as his son (x+10=3(y+10))
or 10+x=3(y+10)

so our sentences are
x=7y
and
x+10=3(y+10)
first we subsitute x=7y for x in the second equation
(7y)+10=3(y+10)
7y+10=3(y+10)
we distribute
7y+10=3y+30
subtract 3y from both sides
4y+10=30
subtract 10 from both sides
4y=20
divide both sides by 4
y=5

the son is currently 5 years old
x=7y
subsitute y=5 for y in equation
x=7(5)
x=35
the dad is currently 35 years old
in 10 years
dad is 45 and son is 15

dad=35
son=5

Answer 2

Answer: these are the answers
son = 5
dad= 35

Related Questions

At a cost of $2.50 per square yard, what would be the price of carpeting a rectangular ﬂoor, 18 ft x 24 ft?

Answers

Answer is: $120

(18ft x 24ft) = 432 square feet

1 yard = 3 feet
1 square yard = (1yd * 1yd) =(3ft * 3ft) = 9 square feet

(432 ft^2)/(9 ft^2 per yard) = 48 square yards

(48 square yards)*($2.50 per yard) = $120

A skateboard company tracks the sales of their short boards and long boards. The company's records show the sales of short boards decrease every three months as represented by expression A, where t is the number of years that the boards have been for sale.expression a:624(o.94)^4t
The company's records show that the sales of long boards increase every four months as represented by expression B, where t is the number of years that the boards have been for sale.
expression b :725(1.12)^3t
Select the statements that give a correct interpretation of the above expressions.

Expression A grows at a rate of 5% every three months, while expression B decays at
a rate of 12% every four months.

Expression A has an initial value of 0.95, while expression B has an initial value of 1.12.

Expression A decays at a rate of 5% every three months, while expression B grows at
a rate of 12% every four months.

Expression A has an initial value of 624, while expression B has an initial value of 725.

Expression A decays at 5% every three years, while expression B grows at a rate of 12% every four years.

Answers

Answer:

Expression A decays at a rate of 5% every three months, while expression B grows at rate of 12% every four months

Expression A has initial value of 624, while expression B has initial value of 725

Step-by-step explanation:

plato correct

i really need help with this you have to choose all the ones that apply please help

Find the unknown angle measure by solving for the given variable.Answer Choices: 32,48,96,24,36,64

Answers

A triangle is 180°. So you can do:

3.2n + 6.4n + 2.4n = 180 Simplify

12n = 180

n = 15 Now that you know the value of n, you can plug it into each individual angle/equation

∠X = 3.2n plug in 15 for n

∠X = 3.2(15)

∠X = 48°

∠Y = 6.4(15)

∠Y = 96°

∠Z = 2.4(15)

∠Z = 36°

What is the numerical coefficient of '-3z'?

Answers

Answer: -3

Step-by-step explanation:

at a high school, the probability that a student is a senior is 0.25. the probability that a student plays a sport is 0.20. the probability that a student is a senior and plays a sport is 0.08. what is the probability that a randomly selected student plays a sport, given that the student is a senior?

Answers

Answer:

0.32

Step-by-step explanation:

We have been given that at a high school, the probability that a student is a senior is 0.25. The probability that a student plays a sport is 0.20. The probability that a student is a senior and plays a sport is 0.08.

We will use conditional probability formula to solve our given problem. $P(B|A)=\frac{P(\text{A and B)}}{P(A)}$ , where,

$P(B|A)$ = The probability of event B given event A.

$P(\text{A and B)}$ = The probability of event A and event B.

$P(A)$ =Probability of event A.

Let A be that the student is senior and B be the student plays a sport.

P(A and B) = Probability that student is a senior and plays a sport.

$P(B|A)=\frac{\text{Probability that a student is a senior and plays a sport}}{\text{Probability that a student is senior}}$

Upon substituting our given values we will get,

$P(B|A)=(0.08)/(0.25)$

$P(B|A)=0.32$

Therefore, the probability that a randomly selected student plays a sport, given that the student is a senior will be 0.32.

How do i solve divide 5/8 by 50%

Answers

50% is $(50)/(100)$
We have two ways to do this, pick your favor:
either by fractions:
$(5)/(8)$ ÷ $(1)/(2)$
( $(50)/(100) = (1)/(2)$
Multiply by the reciprocal of $(1)/(2)$ which is 2(flip it)

$(5)/(8)$ × 2 = $(5)/(4)$ = 1.25

Or change them to decimals:
$(5)/(8) = 0.625$
$(50)/(100)$ = 0.5

0.625 ÷ 0.5

0.5 I 0.625
multiply by 10 to make it easier to divide(it still gives us the same answer)
0.5 × 10 = 5
0.625 × 10 = 6.25
1.25
5 I 6.25
-5
12
-10
25
-25
0

0.625 ÷ 0.5 = 1.25

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