What must be the value of x so that lines a and b are<br> parallel lines cut by transversal f?

Simplify fully, does anyone know ?

Julli [10]

Here's an explanation! :)

6 0

3 years ago

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A baker charges $30 for each batch of cupcakes she makes. She adds $0.09 for each frosting flower. Write an expression to repres

Lisa [10]

Answer:

$30+0.09x$

Here, $x$ denotes number of frosting flowers

Step-by-step explanation:

Let $x$ denotes number of frosting flowers.

Amount charged by baker for each batch of cupcakes = $30

Amount charged for each frosting flower = $0.09

So,

Amount charged for $x$ frosting flowers = $\$0.09x$

Therefore,

Total cost of a batch of cupcakes with frosting flowers = $30+0.09x$

8 0

3 years ago

I can’t figure out how to solve this for a. I’ve tried multiple ways but still have not come up with an answer.

abruzzese [7]

Answer:

a = 33

Step-by-step explanation:

IN a parallelogram consecutive angles are supplementary, thus

5a - 52 + 5a - 98 = 180, that is

10a - 150 = 180 ( add 150 to both sides )

10a = 330 ( divide both sides by 10 )

a = 33

8 0

3 years ago

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Find the inverse of the function.<br> y = 4x

Mashcka [7]

Answer: f-1 (x) = x/4

Step-by-step explanation:

8 0

3 years ago

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

3 years ago

What must be the value of x so that lines a and b are parallel lines cut by transversal f?