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shutvik [7]
3 years ago
8

an arithmetic series has first term 160 and common difference d . the sum of the first 25 terms of the series 3500 . find the co

mmon difference d.
Mathematics
1 answer:
mina [271]3 years ago
6 0

Answer:

d = - \frac{5}{3}

Step-by-step explanation:

The sum to n terms of an arithmetic series is

S_n} = \frac{n}{2} [ 2a₁ + (n - 1)d ]

where a₁ is the first term and d the common difference

Here a₁ = 160, n = 25 and S_{25} = 3500 , thus

\frac{25}{2} [ (2 × 160) + 24d ] = 3500, that is

12.5(320 + 24d) = 3500 ( divide both sides by 12.5 )

320 + 24d = 280 ( subtract 320 from both sides )

24d = - 40 ( divide both sides by 24 )

d = - \frac{40}{24}  = - \frac{5}{3}

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Write an equation for a rational function with: Vertical asymptotes at x = -3 and x = -4 x-intercepts at x = 3 and x = 4 Horizon
sattari [20]

Answer:

(10x ^ 2 - 70 * x +120) / (x ^ 2 + 7 * x +12)

Step-by-step explanation:

First a rational function is a function that is in the form a / b.

Knowing this, we proceed to calculate the asymptotes in x.

Vertical asymptotes

So that there is an asymptote in x, in a rational function the easiest thing is that the denominator is 0. Therefore, for the asymptotes x = -3 and x = -4, we make it 0 in both cases. That is to say:

x + 3 = 0

x + 4 = 0

both, in the denominator. For the moment it would be something like this:

 1 / (x + 3) * (x + 4)

X Intersections

Now, for intersections, what we should do equation 0, and that the solution of x us of the value we want, in this case x = 3 and x = 4. Because here the denominator does not influence because when equal to 0 to a rational function, the denominator becomes 0. Similarly, it is equal to 0 as in the previous case, only this time they would go in the numerator. It would be:

x - 3 = 0

x - 4 = 0

 Therefore, the function would go like this: (x - 3) * (x - 4) / (x +3) * (x + 4),

Operating the following we have:

(x ^ 2 -4 * x - 3 * x +12) / (x ^ 2 + 4 * x + 3 * x +12)

(x ^ 2 - 7 * x +12) / (x ^ 2 + 7 * x +12)

having asymptotes at x = -3 and x = -4, and intersection at (3.0) and (4.0), that is x = 3 and x = 4.

Horizontal asymptote.

Finally, in the case of the horizontal asymptote, which must be y = 10, what must be done is to divide all the terms by the highest degree term, in this case it is x ^ 2. It would be as follows:

(x ^ 2 / x ^ 2 - 7 * x / x ^ 2 + 12 / x ^ 2) / (x ^ 2 / x ^ 2 + 7 * x / x ^ 2 + 12 / x ^ 2)

Now, x ^ 2 / x ^ 2 = 1, and all the others have to be 0, therefore it would remain (1 + 0 + 0) / (1 + 0 + 0) = 1

So for the asymptote to be 10, the entire denominator must be multiplied by 10, thus being:

10 * (x ^ 2 - 7 * x +12) / (x ^ 2 + 7 * x +12)

(10x ^ 2 - 70 * x +120) / (x ^ 2 + 7 * x +12)

And this equation would already comply with everything required in the problem.

8 0
3 years ago
I came up the answer as 57. I will attach my note, can you check?
ololo11 [35]

BC=19

Explanation

Step 1

ABE

triangle ABE is rigth triangle, then let

\begin{gathered} Angle=60 \\ adjacentside=BE \\ opposit\text{ side(the one in front of the angle)= AB=}\frac{19\sqrt[]{6}}{4} \end{gathered}

so, we need a function that relates, angle, adjancent side and opposite side

\tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}}

replace

\begin{gathered} \tan \theta=\frac{opposite\text{ side}}{\text{adjacent side}} \\ \tan 60=\frac{AB}{\text{BE}} \\ \text{cross multiply} \\ \text{BE}\cdot\tan \text{ 60=AB} \\ \text{divide both sides by tan 60} \\ \frac{\text{BE}\cdot\tan\text{ 60}}{\tan\text{ 60}}=\frac{\text{AB}}{\tan\text{ 60}} \\ BE=\frac{\text{AB}}{\tan\text{ 60}} \\ \text{if AB=}\frac{19\sqrt[]{6}}{4} \\ BE=\frac{\frac{19\sqrt[]{6}}{4}}{\sqrt[]{3}} \\ BE=\frac{19\sqrt[]{6}}{4\sqrt[]{3}} \end{gathered}

Step 2

BED

again, we have a rigth triangle,then let

\begin{gathered} \text{Hypotenuse}=BD \\ \text{adjacent side= BE=6.71} \\ \text{angle}=\text{ 45} \end{gathered}

so, we need a function that relates; angle, hypotenuse and adjacent side

\cos \theta=\frac{adjacent\text{ side}}{\text{hypotenuse}}

replace.

\begin{gathered} \cos \theta=\frac{adjacent\text{ side}}{\text{hypotenuse}} \\ \cos 45=\frac{6.71}{\text{BD}} \\ BD=\frac{6.71}{\cos \text{ 45}} \\ BD=\frac{\frac{19\sqrt[]{6}}{4\sqrt[]{3}}}{\frac{\sqrt[]{2}}{2}} \\ BD=\frac{38\sqrt[]{6}}{4\sqrt[]{6}} \\ BD=\frac{38}{4} \end{gathered}

Step 3

finally BDE

let

angle=30

opposite side= BD

use sin function

\begin{gathered} \sin \theta=\frac{opposite\text{ side}}{\text{hypotenuse}} \\ \text{replace} \\ \sin \text{ 30=}\frac{BD}{BC} \\ BC\cdot\sin 30=BD \\ BC=\frac{BD}{\sin \text{ 30}} \\ BC=\frac{\frac{38}{4}}{\frac{1}{2}} \\ BC=\frac{76}{4}=19 \\ BC=19 \end{gathered}

so, the answer is 19

I hop

4 0
1 year ago
Joey practices the guitar every 11th day and the trombone every 3rd day. Joey practiced both the guitar and the trombone today.
densk [106]

11 = 11*1

3=3*1

least common multiple = 3*11*1 = 33

33 days from today he will practice the guitar and the trombone

6 0
3 years ago
In the diagram below, AB is parallel to CD. What is the value of y?
valentinak56 [21]

Answer:

72

Step-by-step explanation:

Corresponding angles on parallel lines always have the same measure.

5 0
2 years ago
It takes 50 minutes to complete 2 math assignments. How long will it take to complete 5 math assignments?
Natalija [7]

Answer:

125 minutes

Step-by-step explanation:

one paper takes 25 minutes to complete. Now if we multiply that by 5 papers you get 125 minutes

7 0
2 years ago
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