All categories

3 years ago

What is the answer to f/2 =5/8

Mathematics

2 answers:

coldgirl [10]3 years ago

8 0

F=2*5/8 which is 5/4

Brut [27]3 years ago

8 0

F is equal to 1.25 because if 5/8 were to be equivalent to f/2, then divide 5/8, both the numerator and the denominator by 4 to get the denominator 2 and the numerator 1.25.

You might be interested in

In the tax schedule below, how much responsibility does a person married filing jointly with an income of $185,356

Keith_Richards [23]

Grub hub gives you perks on the food you love the kinda deals that make you boogie

6 0

3 years ago

What is a perfect circle

Anon25 [30]

Answer:

A circle which has the exact same radius no matter what direction you measure in

Step-by-step explanation:

Probably

6 0

3 years ago

Read 2 more answers

PLS I NEED HELP FAST I DONT UNDER STAND THIS

Delvig [45]

The answer is A dijddndndnbdbdbf

5 0

3 years ago

Use synthetic division to solve (x4 – 1) ÷ (x – 1). What is the quotient?

stepan [7]

For this case, we must build a quotient that, when multiplied by the divisor, eliminates the terms of the dividend until it reaches the rest.

The attached figure shows the quotient given by:

$x ^ 3 + x ^ 2 + x + 1$

Answer:

Quotient: $x ^ 3 + x ^ 2 + x + 1$

See attached image

5 0

3 years ago

Read 2 more answers

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

4 years ago