ABCD is a rectangle, if DB=52 and DC=48, what is the length of BC

Is the following conditional true?<br> If x < 4, then x < 10<br> O yes<br> O no

Kay [80]

Answer: No, i think so.

7 0

1 year ago

If Jesse wants to buy a $75,000 10-year term life insurance policy, and the annual premium rate (per $1000 of face value) for hi

andriy [413]

Your answer is a 175.50

5 0

4 years ago

Read 2 more answers

Christine must buy at least 45 fluid ounces of milk at the store. The store only sells milk in 200 milliliter bottles. If there

ratelena [41]

Since there are 1000 milliliters in a liter, that means that there are 33.8/1000 fluid ouncers per milliliter. That means that each 200 mil bottle is 6.76 fluid ounces or 33.8/1000*200. 45/6.76 = 6.66 but you have to round up to a whole number of bottles since you can buy 2/3 of a bottle which makes your answer 7 bottles.

8 0

3 years ago

Instructions: Given one form of a sequence, write the other form.

Rufina [12.5K]

Answer:

The explicit formula is 15 + (n-1) -4

Step-by-step explanation:

The formula for an explicit formula is an = First number of the sequence (15) + (n-1)(times the amount of integers separating two different sequential numbers) which is -4.

For instance, since 15 is the first number, it will be a1 or 15. Next will be + (n-1) which is the same for every explicit sequence and the multiplier will be the difference between each number, which is -4 in the recursive sequence. As a result, the answer is 15 + (n-1) -4.

7 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

4 years ago