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strojnjashka [21]

3 years ago

9

PQRS is a rectangle.

1 answer:

Anuta_ua [19.1K]3 years ago

8 0

Answer:

diagram pls

then I can answer this question

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Write an equation in slope-intercept form, y=mx+b, for the lines with the following information:

Gekata [30.6K]

Slope intercept formula: y = mx + b where m equals slope and b equals the y-intercept.

Answer: y = -6x + 3

7 0

4 years ago

A submarine descended 500 feet below sea level. It then descended another 275 feet. Write and then evaluate a subtraction expres

kaheart [24]

Answer:

-500-275= -775

or

-500+-275=-775

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

How many edges does a dodecahedron have ?

balu736 [363]

Answer:

12

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please help with these math questions

marta [7]

1. Slope intercept form is y=mx+b, where m is slope and b is y-intercept.
It gave you point (8,12) and slope of -2.
So you already know the value of m, so so far it's y = -2x + b.
Plug the coordinates as x and y: 8 for x and 12 for y.
12 = -2(8) + b
12 = -16 + b
28 = b

Adding b in, you get y = -2x + 28

2. Find the slope, then use point-slope form.
Slope is change in y over change in x, or (y2-y1)/(x2-x1).
m = (9-1)/(2-3) = 8/(-1) = -8

Point slope form: y - y1 = m(x - x1)
y - y1 = -8(x - x1)
Now plug in either of the coordinates as x1 and y1, let's do (3,1).
y - 1 = -8(x - 3)
y - 1 = -8x + 24
y = -8x + 25

3. Do it the same way as number 2. Choose two points, for example (1,1) and (2,-2). You find m = -3 and get final answer as y=-3x + 4.

4. B = 150 + 80d
B is account balance, d is number of monthly deposits.
If you rearrange the terms: B = 80d + 150, you see that that is in slope-intercept form y = mx+b, and if you remember from #1 m is the slope. In this case, the variables are changed, so d is the slope here. And the question told you that d, the slope, is the number of monthly deposits. Its coefficient is 80, so Bianca deposits $80/month.

<span>5. </span>
14a + 16b = 560
16b = 560 - 14a
b = (560 - 14a)/16
b = 560/16 - 14a/16
b = 35 - (7/8)a
<span>b = -(7/8)a + 35</span>

3 0

3 years ago

Sum to n terms of each of following series. (a) 1 - 7a + 13a ^ 2 - 19a ^ 3+...

julia-pushkina [17]

Notice that the difference in the absolute values of consecutive coefficients is constant:

|-7| - 1 = 6

13 - |-7| = 6

|-19| - 13 = 6

and so on. This means the coefficients in the given series

$\displaystyle \sum_{i=1}^\infty c_i a^{i-1} = \sum_{i=1}^\infty |c_i| (-a)^{i-1} = 1 - 7a + 13a^2 - 19a^3 + \cdots$

occur in arithmetic progression; in particular, we have first value $c_1 = 1$ and for $n>1$ , $|c_i|=|c_{i-1}|+6$ . Solving this recurrence, we end up with

$|c_i| = |c_1| + 6(i-1) \implies |c_i| = 6i - 5$

So, the sum to $n$ terms of this series is

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \underbrace{\sum_{i=1}^n i (-a)^{i-1}}_{S'} - 5 \underbrace{\sum_{i=1}^n (-a)^{i-1}}_S$

The second sum $S$ is a standard geometric series, which is easy to compute:

$S = 1 - a + a^2 - a^3 + \cdots + (-a)^{n-1}$

Multiply both sides by $-a$ :

$-aS = -a + a^2 - a^3 + a^4 - \cdots + (-a)^n$

Subtract this from $S$ to eliminate the intermediate terms to end up with

$S - (-aS) = 1 - (-a)^n \implies (1-(-a)) S = 1 - (-a)^n \implies S = \dfrac{1 - (-a)^n}{1 + a}$

The first sum $S'$ can be handled with simple algebraic manipulation.

$S' = \displaystyle \sum_{i=1}^n i (-a)^{i-1}$

$\displaystyle S' = \sum_{i=0}^{n-1} (i+1) (-a)^i$

$\displaystyle S' = \sum_{i=0}^{n-1} i (-a)^i + \sum_{i=0}^{n-1} (-a)^i$

$\displaystyle S' = \sum_{i=1}^{n-1} i (-a)^i + \sum_{i=1}^n (-a)^{i-1}$

$\displaystyle S' = \sum_{i=1}^n i (-a)^i - n (-a)^n + S$

$\displaystyle S' = -a \sum_{i=1}^n i (-a)^{i-1} - n (-a)^n + S$

$\displaystyle S' = -a S' - n (-a)^n + \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle (1 + a) S' = \dfrac{1 - (-a)^n - n (1 + a) (-a)^n}{1 + a}$

$\displaystyle S' = \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2}$

Putting everything together, we have

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 S' - 5 S$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2} - 5 \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} =\boxed{\dfrac{1 - 5a - (6n+1) (-a)^n + (6n-5) (-a)^{n+1}}{(1+a)^2}}$

8 0

2 years ago