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Nutka1998 [239]
3 years ago
12

What is the value of E(-1) n (3n+2)?

Mathematics
1 answer:
RoseWind [281]3 years ago
6 0
<h3>Answer:  C) 6</h3>

====================================================

Explanation:

The weird looking E symbol is the greek uppercase letter sigma. It refers to a sum.

It tells us to add up terms in the form (-1)^n*(3n+2) where n is an integer ranging from n = 1 to n = 4.

------------------

If n = 1, then we have

(-1)^n*(3n+2) = (-1)^1*(3*1+2) = -5

Let A = -5 as we'll use it later.

------------------

If n = 2, then

(-1)^n*(3n+2) = (-1)^2*(3*2+2) = 8

Let B = 8 since we'll use this later as well

------------------

If n = 3, then

(-1)^n*(3n+2) = (-1)^3*(3*3+2) = -11

Let C = -11

-------------------

If n = 4, then

(-1)^n*(3n+2) = (-1)^4*(3*4+2) = 14

Let D = 14.

--------------------

We'll add up the values of A,B,C,D to get the final answer

A+B+C+D = -5+8+(-11)+14 = 6

This means that

\displaystyle \sum_{n=1}^{4}(-1)^n(3n+2) = 6

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4 years ago
13. For what value of b would the following system of equations have an infinite number of solutions?
tiny-mole [99]

Answer:

b=2

Step-by-step explanation:

we have

9x+12y=21 -----> equation A

6x+8y=7b ----> equation B

we know that

If the system of equations have an infinite number of solutions then the equation A must be equal to the equation B

Multiply equation B by 1.5 both sides

1.5*[6x+8y[=7b*1.5

9x+12y=10.5b ----> equation C

Compare equation A and equation C

9x+12y=21 -----> equation A

9x+12y=10.5b ----> equation C

For the equations to be equal it must be fulfilled that

21=10.5b

solve for b

b=21/10.5

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5 0
4 years ago
Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

y=ax^2+bx+c            ...(i)

It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

The quadratic function passes through the point (5,31). So, substitute x=5,y=31 in (ii).

31=a(5)^2+b(5)+11

31-11=a(25)+5b

20=25a+5b

Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

11-11=a(9)+3b

0=9a+3b

Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

4-0=5a+b-3a-b

4=2a

\dfrac{4}{2}=a

2=a

Putting a=2 in (iv), we get

0=3(2)+b

0=6+b

-6=b

Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

7 0
3 years ago
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