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2 years ago

7

The dad has a son and a daughter. The daughter is 19 years younger than the son. The son is 13 years younger than the dad (Don't

ask, just keep reading.) The daughter is 21 years old now. How old is the dad?
Dad - 13 years older than son
Son - 19 years older than daughter
Daughter - 21 years old

2 answers:

Ivenika [448]2 years ago

6 0

Answer:

53

Step-by-step explanation:

Elis [28]2 years ago

5 0

Answer:

The Dad is 53

Step-by-step explanation:

21 + 19 = 40 ( Brothers age )

40 + 13 = 53 (Dads age )

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3 years ago

The length of a swimming pool is 3 feet longer than it’s width. The swimming pool is surrounded by a deck that is 2 feet wide an

Firlakuza [10]

Answer:

The Length of swimming pool is 8.35 feet

The width of swimming pool is 5.35 feet

Step-by-step explanation:

Given as :

The length of swimming pool is 3 feet longer that its width

Let The width of swimming pool = w feet

So, The Length of swimming pool = L = (w + 3) feet

Now, The swimming pool is surrounded by deck of 2 feet wide

The width of deck = w' = w + 2 + 2 = (w + 4) feet

The area of deck = 116 feet²

The length of deck = L' = L + 2 + 2 = (L + 4) feet

So, L' = (w + 3 + 4) feet

I.e L' = (w + 7) feet

∵ The area of deck = 116 feet²

So , L' × w' = 116 feet²

(w + 7)× (w + 4) = 116

Or, w² + 4 w + 7 w + 28 = 116

Or, w² + 11 w - 88 = 0

Solving the quadratic equation as

ax² + bx + c = 0

So, w = $\dfrac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}$

Or, w = $\dfrac{-11\pm \sqrt{11^{2}-4\times 1\times (-88)}}{2\times 1}$

Or, w = $\dfrac{-11\pm \sqrt{473}}{2}$

or, w = $\dfrac{-11\pm 21.7}{2}$

or, w = 5.35 , -16.35

The width of swimming pool = w = 5.35 feet

The Length of swimming pool = L = (5.35 + 3) feet = 8.35 feet

Hence, The Length of swimming pool is 8.35 feet

And The width of swimming pool is 5.35 feet

Answer

3 0

3 years ago

Perry borrows $350 at a simple interest rate of 4.5%. Perry pays back the loan in 30 months. How much interest does Perry pay on

Nitella [24]

Answer: the interest on the loan is $39.38

Step-by-step explanation:

The formula for determining simple interest is expressed as

I = PRT/100

Where

I represents interest paid on the loan.

P represents the principal or amount taken as loan

R represents interest rate

T represents the duration of the loan in years.

From the information given,

P = $350

R = 4.5%

There are 12 months in a year. Converting 30 months into years, it becomes

30/12 = 2.5. so

T = 2.5 years

Therefore

I = (350 × 4.5 × 2.5)/100

I = $39.38

3 0

3 years ago

Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle

Anastasy [175]

Answer:

Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x+3)^{2} +(y-4)^{2}=r^{2}$

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

$r=\sqrt{(2-4)^{2}+(-6+3)^{2}}$

$r=\sqrt{(-2)^{2}+(-3)^{2}}$

$r=\sqrt{13}\ units$

The equation of the circle is equal to

$(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}$

$(x+3)^{2} +(y-4)^{2}=13$

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

$(10+3)^{2} +(4-4)^{2}=13$

$(13)^{2} +(0)^{2}=13$

$169=13$ -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

$(x-1)^{2} +(y-3)^{2}=r^{2}$

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

$r=\sqrt{(6-3)^{2}+(2-1)^{2}}$

$r=\sqrt{(3)^{2}+(1)^{2}}$

$r=\sqrt{10}\ units$

The equation of the circle is equal to

$(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}$

$(x-1)^{2} +(y-3)^{2}=10$

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

$(11-1)^{2} +(5-3)^{2}=10$

$(10)^{2} +(2)^{2}=10$

$104=10$ -----> is not true

therefore

The point is not on the circle

The statement is false

7 0

3 years ago