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3 years ago

Which term of series1 +4+7+10+...is43

Mathematics

1 answer:

Jlenok [28]3 years ago

4 0

Answer:

a = 4

d = a2 - a1

d = 7 - 4

d = 3

an = 43

an = a + (n - 1)d

43 = 4 + (n - 1) 3

43 = 4 + 3n - 3

43 = 1 + 3n

43 - 1 = 3n

3n = 42

n = 42 ÷

n = 14

So, 14th term is 43

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Doss [256]

The median is the number that is in the middle of all the other numbers in order but that is not the case
however, it is not the mean either
so we can see that 9.25 is between 9 and 9.5 when 23.5 is ignored
the correct answer is B

8 0

4 years ago

Is 9 a solution to the equation 21=3p-5? Show work

Dmitry_Shevchenko [17]

Answer:

Step-by-step explanation:

21 = 3p - 5

<u>Step 1</u> : Add 5 on both sides

21 + 5 = 3p

26 = 3p

3p = 26

<u>Step 2</u> : Divide 3 on both sides

p = 26/3

Hence, the solution of the given equation is 26/3, not 9.

4 0

2 years ago

Identify the volume and surface area of the sphere in terms of π.

Anastasy [175]

Answer:

Sphere Formulas in terms of radius r:

Volume of a sphere: V = (4/3)πr.

5 0

3 years ago

Unsure how to do this calculus, the book isn't explaining it well. Thanks

krok68 [10]

One way to capture the domain of integration is with the set

$D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}$

Then we can write the double integral as the iterated integral

$\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx$

Compute the integral with respect to $y$ .

$\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)$

Compute the remaining integral.

$\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}$

We could also swap the order of integration variables by writing

$D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}$

and

$\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy$

and this would have led to the same result.

$\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)$

$\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)$

7 0

1 year ago

What is the equation of the following line? Be sure to scroll down first to see

Rom4ik [11]

<h2>Answer:</h2>

-1/3x

<h2>Step-by-step explanation:</h2><h3>Known :</h3>

x1 = -3
x2 = 0
y1 = 1
y2 = 0

<h3>Asked :</h3>

Correct line equation

<h3>Solution :</h3>

We can find the straight line equation using this formula,

$\frac{y-y1}{y2-y1} = \frac{x-x1}{x2-x1}$

Insert all we know inside the formula,

$\frac{y-1}{0-1} = \frac{x-(-3)}{0-(-3)}\\\frac{y-1}{-1} = \frac{x +3}{3}\\$

Use cross multiplication,

3(y - 1) = -1(x + 3)\

3y - 3 = -x - 3

3y = -x - 3 + 3

3y = -x

y = -1/3x

<h3>Conclusion :</h3>

The correct equation is y = -1/3x

6 0

3 years ago