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3 years ago

The table shows pattern p and patterns q complete the table

Mathematics

2 answers:

jek_recluse [69]3 years ago

7 0

Answer:P=8 and Q=16

Step-by-step explanation:

mario62 [17]3 years ago

5 0

ANSWER

P=8,Q=16

EXPLANATION

We can see from the table that both pattern P and Q have an initial value of 24.

Pattern P is reducing by a constant difference of 4.

The next value for pattern P is

12-4=8.

Also pattern Q is reducing by a constant value of 2.

The next value for pattern Q is,

18-2=16

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Anastaziya [24]

Answer: 20% = 0.2

Step-by-step explanation:

-Add all the ball for a total frequency (15 balls)

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5/15= 0.33

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-multiply by 100 to get a percentage.

8 0

3 years ago

Read 2 more answers

Harry has $25 in her savings account she will add $5 each week and not take any money out of her account which expression repres

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$5 X 52( weeks in year) +$25= y

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3 years ago

Please help with my geometry

erma4kov [3.2K]

25 = hypothenuse
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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!

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4 years ago

If the farmer has 276 feet of fencing, what are the dimensions of the region which enclose the maximal area?

gtnhenbr [62]

Answer:

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Step-by-step explanation:

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3 years ago