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Mice21 [21]
3 years ago
15

For the following right triangle find the side length x round your answer to the nearest hundredth

Mathematics
1 answer:
pashok25 [27]3 years ago
3 0

Answer:

x ≈ 12.96

Step-by-step explanation:

Using Pythagoras' identity in the right triangle

x² + 11² = 17²

x² + 121 = 289 ( subtract 121 from both sides )

x² = 168 ( take the square root of both sides )

x = \sqrt{168} ≈ 12.96 ( to the nearest hundredth )

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The equation |2x-8|+3= 5 has 2 solutions. These
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Answer:  x=8  x=0

Step-by-step explanation:

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3 years ago
Whet is 2y+14x-7x+9y
matrenka [14]

Answer:

Step-by-step explanation:

2y + 14x - 7x + 9y

2y + 9y + 14x - 7x

11y + 7x

6 0
3 years ago
Read 2 more answers
How do you do this question?
likoan [24]

Answer:

a) -7/9

b) 16 / (n² + 15n + 56)

c) 1

Step-by-step explanation:

When n = 1, there is only one term in the series, so a₁ = s₁.

a₁ = (1 − 8) / (1 + 8)

a₁ = -7/9

The sum of the first n terms is equal to the sum of the first n−1 terms plus the nth term.

sₙ = sₙ₋₁ + aₙ

(n − 8) / (n + 8) = (n − 1 − 8) / (n − 1 + 8) + aₙ

(n − 8) / (n + 8) = (n − 9) / (n + 7) + aₙ

aₙ = (n − 8) / (n + 8) − (n − 9) / (n + 7)

If you wish, you can simplify by finding the common denominator.

aₙ = [(n − 8) (n + 7) − (n − 9) (n + 8)] / [(n + 8) (n + 7)]

aₙ = [n² − n − 56 − (n² − n − 72)] / (n² + 15n + 56)

aₙ = 16 / (n² + 15n + 56)

The infinite sum is:

∑₁°° aₙ = lim(n→∞) sₙ

∑₁°° aₙ = lim(n→∞) (n − 8) / (n + 8)

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8 0
3 years ago
1. Which of the following is an arithmetic sequence?
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i say C. -5, -2, 1, 4, 7 is your answer for question 1.

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7 0
3 years ago
Read 2 more answers
X=25.36+0.45(25.36)<br><br> LaTeX: x=
e-lub [12.9K]

Answer:

x=36.772  

Step-by-step explanation:

Given : Expression x=25.36+0.45(25.36)

To find : Solve the expression ?

Solution :

Step 1 - Write the expression,

x=1(25.36)+0.45(25.36)

Step 2 - Apply distributive, ab+ac=a(b+c)

Here, a=25.36, b=1, c=0.45

x=25.36(1+0.45)

Step 3 - Solve the expression,

x=25.36\times 1.45

x=36.772

Therefore, x=36.772

7 0
3 years ago
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