Write in slop intercept form

Charlie had a full bottle of water. He drank 1/4 of the water in the morning. He drank 2/5 of the water with lunch. How much of

12345 [234]

Answer:

7/20

Step-by-step explanation:

first of all you have to put the two fractions on the same denominators

1/4 (multiply by 5) and you will get 5/20

2/5(multiply by 4) and you will get 8/20

we have to add 5/20 and 8/20 that gives us 13/20

13/20 is the amount he drank

20/20 - 13/20 =7/20

7/20 is the amount of water left that Charlie has left

3 0

3 years ago

Nolan's car used 2 gallons to travel 50 miles. How far can he travel on 3 gallons?

ElenaW [278]

Answer:

On 3 gallons, he could travel 75 miles

Step-by-step explanation:

50 / 2 = 25 miles per gallon

25 * 3 = 75miles

5 0

3 years ago

According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the

FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let $P(t)$ be the amount of $^{235}U$ and $Q(t)$ be the amount of $^{238}U$ after $t$ years.

Then, we obtain two differential equations

$\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q$

where $k_1$ and $k_2$ are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

$\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt$

Now, the variables are separated, $P$ and $Q$ appear only on the left, and $t$ appears only on the right, so that we can integrate both sides.

$\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt$

which yields

$\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2$ ,

where $c_1$ and $c_2$ are constants of integration.

By taking exponents, we obtain

$e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}$

Hence,

$P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}$ ,

where $C_1 := \pm e^{c_1}$ and $C_2 := \pm e^{c_2}$ .

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

$P(0) = Q(0) = C$

Substituting 0 for $P$ in the general solution gives

$C = P(0) = C_1 e^0 \implies C= C_1$

Similarly, we obtain $C = C_2$ and

$P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}$

The relation between the decay constant $k$ and the half-life is given by

$\tau = \frac{\ln 2}{k}$

We can use this fact to determine the numeric values of the decay constants $k_1$ and $k_2$ . Thus,

$4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}$

and

$7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}$

Therefore,

$P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}$

We have that

$\frac{P(t)}{Q(t)} = 137.7$

Hence,

$\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7$

Solving for $t$ yields $t \approx 6 \times 10^9$ , which means that the age of the universe is about 6 billion years.

5 0

3 years ago

HELP PLEASE!! WILL MARK BRAINLIEST!!!!

elixir [45]

Answer:

see below

Step-by-step explanation:

$\frac{1}{x^{2} } + \frac{2}{y}/ \frac{5}{x} -\frac{6}{y^2}$

find common denominator and add the fractions in numerator and denominator:

1/x² + 2/y = y+2x²/x²y

5/x -6/y² =5y² -6x/xy²

(y+2x²)/x²y / (5y²-6x)/xy² change into multiplication

(y+2x²)/x²y * xy²/(5y²-6x)

$\frac{y+2x^{2} }{x^{2} y} *\frac{xy^{2} }{5y^{2}-6x }$ simplify

y(y+2x²)/ x(5y²-6x)

4 0

3 years ago

Formula area of the square

Misha Larkins [42]

A= l * W (area = length times width)

4 0

3 years ago

Write in slop intercept form​

Write in slop intercept form