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Nadusha1986 [10]
2 years ago
11

4) An arithmetic sequence has a 7th term of 54 and a 13th term of 94. Find the common difference

Mathematics
1 answer:
san4es73 [151]2 years ago
6 0

Answer:

20/3

Step-by-step explanation:

Every nth term takes the form of  a + n*d, where a is the first  term.

So 7th term = (54 = a + 7d), 13th term = (94 = a + 13d).

equate them both:

94 - 13d = 54 - 7d

40 = 6d

d = 40/6

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During the 2005 football season team tackle the team sack by eight points if the combine scores total of 26 find the individual
Ierofanga [76]

Answer:

Team sack = 9

Team tackle = 17

Step-by-step explanation:

To start with, we'd assume that

Let x = the points scored by Team Sack

Let x + 8 = the points scored by Team Tackle

x + x + 8 = 26

So, go further and solve mathematically

2x + 8 = 26

if we subtract 8 from both sides, we have

2x = 18

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x = 18/2

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2 years ago
Drag each length to the correct location on the image. Each length can be used more than once, but not all lengths will be used.
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Applying trigonometric ratios, the missing segments lengths in the image attached below are: <em>a = 8√3; b = 12.</em>

<h3>The Trigonometric Ratios</h3>
  • SOH is sin ∅ = opp/hyp
  • CAH is cos ∅ = adj/hyp
  • TOA is tan ∅ = opp/adj
  • SOHCAHTOA is used to solve right triangles.

Given the right triangle in the mage attached below, we would find the missing sides as follows:

<em>Find a:</em>

Reference angle (∅) = 30°

opposite = 4√3

hypotenuse = a

  • <em>Apply SOH:</em>

sin 30 = (4√3)/a

a = (4√3)/sin 30

a = (4√3)/(1/2) (sin 30 = 1/2)

a = (4√3)×2

a = 8√3

<em>Find b:</em>

Reference angle (∅) = 60°

opposite = b

adjacent = 4√3

  • <em>Apply TOA:</em>

tan 60 = b/(4√3)

b = tan 60 × 4√3

b = √3 × 4√3

b = 12

Therefore, applying trigonometric ratios, the missing segments lengths in the image attached below are: <em>a = 8√3; b = 12.</em>

Learn more about trigonometric ratios on:

brainly.com/question/4326804

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1 year ago
Help this question pls I mark pro brainlinest​
omeli [17]
1st option is the answer
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2 years ago
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Solve this identifying holes, vertical asymptotes, and horizontal asymptotes for
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\lim\limits_{x\to\pm\infty}\dfrac{x^2+7x+12}{-x^2-3x+4}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\dfrac{7}{x}+\dfrac{12}{x^2}\right)}{x^2\left(-1-\dfrac{3}{x}+\dfrac{4}{x^2}\right)}=\dfrac{1}{-1}=-1\\\\\boxed{y=-1}

6 0
3 years ago
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