Solve for x. Enter the solutions from least to greatest.

Five years ago, Victor was twice as old as his daughter Anika. In 10 years, the sum of their ages will equal 90. Find their ages

frozen [14]

Answer: 45 and 25

Step-by-step explanation:

Given

Five years ago victor was twice as old as his daughter

Suppose the current age of victor and his daughter are x and y

For five years ago

$\Rightarrow (y-5)=2(x-5)\\\Rightarrow y-5=2x-10\\\Rightarrow y=2x-5\\\Rightarrow 2x-y=5\quad \ldots(i)$

In 10 years sum of their ages is 90

$\Rightarrow (y+10)+(x+10)=90\\\Rightarrow x+y=70\quad \ldots(ii)$

On solving (i) and (ii) we get

$\Rightarrow x=25,y=45$

So, the current age of victor is 45 and his daughter is 25

8 0

2 years ago

There are some Nichols dimes and quarters in a large piggy bank for every two nickels there a sweetheart forever to Diane’s ther

grigory [225]

Answer:

There are 80 nickels, 120 dimes and 300 quarters in the piggy bank.

All coins in the piggy bank worth $91.

Step-by-step explanation:

Please consider the complete question:

There are some nickels, dimes, and quarters in a large piggy bank. For every 2 nickels there are 3 dimes. For every 2 dimes there are 5 quarters. There are 500 coins total. How many nickels, dimes, and quarters are in the piggy bank? Explain your reasoning. How much are the coins in the piggy bank worth?

Let n, d, q represent number of nickles, dimes and quarters respectively.

For every 2 nickels there are 3 dimes. We can represent this information in an equation as:

$\frac{2}{n}=\frac{3}{d}...(1)$ or

$n=\frac{2d}{3}...(1)$

For every 2 dimes there are 5 quarters. We can represent this information in an equation as:

$\frac{2}{d}=\frac{5}{q}...(2)$ or

$q=\frac{5d}{2}...(2)$

There are 500 coins total. We can represent this information in an equation as:

$n+d+q=500...(3)$

We will use substitution method to solve our given system.

Upon substituting equation (1) and equation (2) in equation (3), we will get:

$\frac{2d}{3}+d+\frac{5d}{2}=500$

Multiply the equation by 6:

$\frac{2d}{3}*6+d*6+\frac{5d}{2}*6=500*6$

$2d*2+6d+5d*3=3000$

$4d+6d+15d=3000$

$25d=3000$

$\frac{25d}{25}=\frac{3000}{25}$

$d=120$

Therefore, there are 120 dimes in the piggy bank.

Upon substituting $d=120$ in equation (1), we will get:

$n=\frac{2(120)}{3}$

$n=2(40)$

$n=80$

Therefore, there are 80 nickels in the piggy bank.

Upon substituting $d=120$ in equation (2), we will get:

$q=\frac{5d}{2}$

$q=\frac{5(120)}{2}$

$q=5(60)$

$q=300$

Therefore, there are 300 quarters in the piggy bank.

Now, we need to find the value of all coins.

We know 1 quarter is worth $0.25, so 300 quarters will be worth $\$0.25\times 300=\$75$ .

We know 1 dime is worth $0.10, so 120 dimes will be worth $\$0.10\times 120=\$12$ .

We know 1 nickel is worth $0.05, so 80 nickels will be worth $\$0.05\times 80=\$4$ .

$\text{All coins worth}=\$75+\$12+\$4$

$\text{All coins worth}=\$91$

Therefore, all coins in the piggy bank worth $91.

7 0

3 years ago

The length of some fish are modeled by a von Bertalanffy growth function. For Pacific halibut, this function has the form L(t) =

Dafna1 [17]

Answer:

a) L'(t) = 34.416*e^(-0.18*t)

b) L'(0) = 34 cm/yr , L'(1) =29 cm/yr , L'(6) =12 cm/yr

c) t = 10 year

Step-by-step explanation:

Given:

- The length of fish grows with time. It is modeled by the relation:

L(t) = 200*(1-0.956*e^(-0.18*t))

Where,

L: Is length in centimeter of a fist

t: Is the age of the fish in years.

Find:

(a) Find the rate of change of the length as a function of time

(b) In this part, give you answer to the nearest unit. At what rate is the fish growing at age: t = 0 , t = 1, t = 6

c) When will the fish be growing at a rate of 6 cm/yr? (nearest unit)

Solution:

- The rate of change of length of a fish as it ages each year can be evaluated by taking a derivative of the Length L(t) function with respect to x. As follows:

dL(t)/dt = d(200*(1-0.956*e^(-0.18*t))) / dt

dL(t)/dt = 34.416*e^(-0.18*t)

- Then use the above relation to compute:

L'(t) = 34.416*e^(-0.18*t)

L'(0) = 34.416*e^(-0.18*0) = 34 cm/yr

L'(1) = 34.416*e^(-0.18*1) = 29 cm/yr

L'(6) = 34.416*e^(-0.18*6) = 12 cm/yr

- Next, again use the derived L'(t) to determine the year when fish is growing at a rate of 6 cm/yr:

6 cm/yr = 34.416*e^(-0.18*t)

e^(0.18*t) = 34.416 / 6

0.18*t = Ln(34.416/6)

t = Ln(34.416/6) / 0.18

t = 10 year

7 0

3 years ago

Which statement best describes the area of the triangle shown below? A coordinate grid is shown with a triangle.. The base is 4

drek231 [11]

It willllllllll be B

3 0

3 years ago

Read 2 more answers

BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!

mash [69]

Part 1:

Given that a store had 235 MP3 players in the month of January and that every month, 30% of the MP3 players were sold and 50 new MP3 players were stocked in the store.

The number of MP3 players in the store after the previous months sale is given by 0.7f(n - 1) and the number of MP3 players in the store after new MP3 were added is 0.7f(n - 1) + 50.

Therefore, the recursive function that best represents the number of MP3 players in the store f(n) after n months is given by f(n) = 0.7 x f(n − 1) + 50, f(0) = 235, n > 0

Part 2:

The average rate of change of a function from a to b is given by

$average \ rate \ of \ change= \frac{f(b)-f(a)}{b-a}$

Given that the equation showing the value of her investment after x years is given by

$f(x)=500(1.05)^x$

Thus, the average rate of change of the value of Sophia's investment from the second year to the fourth year is given by

$Average \ rate \ of \ change= \frac{f(4)-f(2)}{4-2} \\ \\ = \frac{500(1.05)^4-500(1.05)^2}{2} = \frac{607.75-551.25}{2} \\ \\ = \frac{56.5}{2} =28.25$

Therefore, the average rate of change of the value of Sophia's investment from the second year to the fourth year is 28.25 dollars per year.

7 0

3 years ago

Read 2 more answers