Answer:
The sum of the first 37 terms of the arithmetic sequence is 2997.
Step-by-step explanation:
Arithmetic sequence concepts:
The general rule of an arithmetic sequence is the following:
In which d is the common diference between each term.
We can expand the general equation to find the nth term from the first, by the following equation:
The sum of the first n terms of an arithmetic sequence is given by:
In this question:
We want the sum of the first 37 terms, so we have to find
Then
The sum of the first 37 terms of the arithmetic sequence is 2997.
Answer:
198 cubic metres
Steps:
<em><u>Find </u></em><em><u>the </u></em><em><u>area </u></em><em><u>of </u></em><em><u> </u></em><em><u>cross</u></em><em><u> </u></em><em><u>section</u></em>
<em><u>Area </u></em><em><u>if </u></em><em><u>traing</u></em><em><u>l</u></em><em><u>e </u></em><em><u>+</u></em><em><u> </u></em><em><u>Area </u></em><em><u>of </u></em><em><u>square</u></em>
<em><u>1</u></em><em><u>/</u></em><em><u>2</u></em><em><u> </u></em><em><u>×</u></em><em><u> </u></em><em><u>3</u></em><em><u> </u></em><em><u>×</u></em><em><u> </u></em><em><u>4</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>4</u></em><em><u>×</u></em><em><u>4</u></em>
<em><u>6</u></em><em><u>+</u></em><em><u> </u></em><em><u>1</u></em><em><u>6</u></em>
<em><u>2</u></em><em><u>2</u></em>
<em><u>Then </u></em><em><u>times </u></em><em><u>with </u></em><em><u>the </u></em><em><u>lenght </u></em><em><u>(</u></em><em><u>9</u></em><em><u>)</u></em>
<em><u>2</u></em><em><u>2</u></em><em><u>×</u></em><em><u>9</u></em>
I hope this helps, dont hesitate to ask for any question.
Mark me as brainliest is appreciated.Tq!!
What is the solution set of x2 + y2 = 26 and x − y = 6? A. {(5, -1), (-5, 1)} B. {(1, 5), (5, 1)} C. {(-1, 5), (1, -5)} D. {(5,
Rus_ich [418]
He two equations given are
x^2 + y^2 = 26
And
x - y = 6
x = y +6
Putting the value of x from the second equation to the first equation, we get
x^2 + y^2 = 26
(y + 6) ^2 + y^2 = 26
y^2 + 12y + 36 + y^2 = 26
2y^2 + 12y + 36 - 26 = 0
2y^2 + 12y + 10 = 0
y^2 + 6y + 5 = 0
y^2 + y + 5y + 5 = 0
y(y + 1) + 5 ( y + 1) = 0
(y + 1)(y + 5) = 0
Then
y + 1 = 0
y = -1
so x - y = 6
x + 1 = 6
x = 5
Or
y + 5 = 0
y = - 5
Again x = 1
So the solutions would be (-1, 5), (1 , -5). The correct option is option "C".