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kipiarov [429]
3 years ago
15

For the Rule of 78, for a 12-month period, the last term in the sequence is 12 and the series sums to 78. For an 10 month period

, the last term is__ and the series sum is__ . For a 15 month period, the last term is __ and the series sum is . For a 20 month period, the last term is__
Mathematics
1 answer:
Svetllana [295]3 years ago
5 0
For a 10 month period the last term is 10 and the series sum is 55. For a 15 month period the last term is 15 and the sum is 120. For a 20 month period the last term is 20.
If u want the sum then it is 210.

Hope this helps you!
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kirill [66]
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2 years ago
The figure shown is a kite. What is the x-coordinate of point P?
tresset_1 [31]

The answer to the above question can be determined as -

Let the point on the left side of P be Q, thus coordinates of Q are (a,1) and point on the right side of P be R, thus coordinates of R will be, (a+4, 1).

Now, it given that, between x coordinates of Q and R is 4, and it can be seen that they are getting divided into half.

So, the x coordinate of P will be - a + \frac{4}{2} i.e. a + 2

<u>Thus, the x coordinate of P will be a +2.</u>

6 0
3 years ago
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If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?
lys-0071 [83]

Answer:

2

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

3x^2-(3k-2)x-(k-6) \text{ with }k=2:

3x^2-(3\cdot 2-2)x-(2-6)

3x^2-4x+4

I'm going to solve 3x^2-4x+4=0 for x using the quadratic formula:

\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}

\frac{4\pm \sqrt{16-16(3)}}{6}

\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}

\frac{4\pm 4\sqrt{-2}}{6}

\frac{2\pm 2\sqrt{-2}}{3}

\frac{2\pm 2i\sqrt{2}}{3}

Let's see if uv=u+v holds.

uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}

Keep in mind you are multiplying conjugates:

uv=\frac{1}{9}(4-4i^2(2))

uv=\frac{1}{9}(4+4(2))

uv=\frac{12}{9}=\frac{4}{3}

Let's see what u+v is now:

u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}

u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}

We have confirmed uv=u+v for k=2.

4 0
2 years ago
Given the linear equation 3x + y = 5, find the slope of its graph
umka2103 [35]

Answer:

B) -3

Step-by-step explanation:

1. First you have to write the equation in slope intercept form. To do that, you must have y be alone on one side.

2. move the 3x to the right side with the 5 by subtracting 3x from both sides. Since 3x will cancel on the left, y will be left alone. y=-3x+5

3. Now that you're in slope intercept form, you can find the slope easily. The slope will always be the number in front of x. In this case, the number in front of x is -3. Therefore, -3 is your slope!

7 0
3 years ago
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Solve the system <br> {f (x) = 2x-1<br> {g (x) = x^2-4
Murrr4er [49]

Answer:

2 sets of possible solutions:

x=3, y = 5

and

x=-1, y = -3

Step-by-step explanation:

Using the graphical method, (see attached)

you can graph both equations and find their intersection points.

From the attached plot, you can see that the graphs intersect at (3,5) and (-1,-3)

Alternatively, you can solve this numerically by solving the following system of equations. You will get the same answer.

y = 2x + 1 ------------------- eq. (1)

y = x² - 4 ------------------- eq. (2)

4 0
3 years ago
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