Find the slope and the y-intercept of the line 6x+2y=5.

Find the sum of the counting numbers from 1 to 25 inclusive. In other words, if S = 1 + 2 + 3 + ... + 24 + 25, find the value of

Anastaziya [24]

Answer:

325

Step-by-step explanation:

You must have heard about Arithmetic Progressions (AP)

Arithmetic progressions are a series of numbers such that every successive number is the sum of a constant number and the previous number.

Our very own counting numbers form AP

For example :-

2 = 1 + <u>1</u>

3 = 2 + <u>1</u>

4 = 3 + <u>1</u>

The number in bold (1) is that constant number which is added to a number to form its successive number.

To find the sum of series forming AP, we use the formula :-

$sum = \frac{n}{2} \{ a + a _{n} \}$

here,

n is the number of terms
a is the first number of the series
an is the last number of the series

So we'll use all this information to find the sum of continuous numbers from 1 to 25 where 1 is the first term(a) and 25 is the last(an).

and n is 25

$S = \frac{25}{2}\{ 1 +25\}$

$= \frac{25 \times 26}{2}$

$= 25 \times 13$

$= 325$

So, the value of S comes out to be 325.

8 0

3 years ago

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Please answer this correctly

Aloiza [94]

Answer:

Height of this missing bar would be 1

Step-by-step explanation:

Since there is 1 and only 1 quantity between 80-99.

5 0

3 years ago

Read 2 more answers

Solve the linear equation. 14x − 1/2 (4x+6) = 3 (x−4) − 18

LekaFEV [45]

$14x - (2x + 3) = 3x - 12 - 18 \\ 14x - 2x - 3 = 3x - 30 \\ 12x - 3 = 3x - 30 \\ 9x = - 27 \\ x = - 3$

7 0

4 years ago

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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

Given 1 in = 2.54 cm and 1 cm = 10 mm, how many millimeters are present in 20.0 inches?

Artemon [7]

Answer: 508mm

Step-by-step explanation:

20 x 2.54 = 50.8

50.8 x 10 = 508

7 0

3 years ago