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Degger [83]
3 years ago
12

A leg of a right triangle is 25 meters long, and the hypotenuse is 35 meters long. What is the length of missing leg?

Mathematics
1 answer:
san4es73 [151]3 years ago
4 0
600 meters hope it help
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Solve the derivative of f(x)= x^4e^tanx
bulgar [2K]

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Step-by-step explanation:

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{RO5-1} Simplify:<br> 2. 73
Volgvan

Answer:

2 11/15

Step-by-step explanation:

Let X equal the decimal number

Equation 1:

X=2.73¯¯¯(1)

With 1 digits in the repeating decimal group,

create a second equation by multiplying

both sides by 101 = 10

Equation 2:

10X=27.33¯¯¯(2)

Subtract equation (1) from equation (2)

10XX9X===27.33...2.73...24.6

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Multiply to eliminate 1 decimal places.

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= 101 = 10

24.69×1010=24690

Find the Greatest Common Factor (GCF) of 246 and 90, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 6,

246÷690÷6=4115

Simplify the improper fraction,

=21115

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5 0
3 years ago
Find the distance between the point (2, 3) and the line 4x + 3y = 10 (round your answer to the nearest tenth).
Monica [59]

Explanation

Let's first graph the elements:

Using the distance formula we get:

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4 0
1 year ago
Line m is parallel to line p. m ∠ HEF = 39º and m ∠ IGF = 13º. Find the m ∠ EFG. Explain in detail how you know you are correct.
svetlana [45]

Answer:

The measure of ∠EFG is 52°

Step-by-step explanation:

Given line m is parallel to line p. m∠HEF = 39º and m∠IGF = 13º.we have to find m∠EFG.

In ΔJFG,

By angle sum property of triangle, which states that sum of all angles of triangle is 180°

m∠FJG+m∠JGF+m∠JFG=180°

⇒ 39°+13°+m∠JFG=180°

⇒ m∠JFG=180°-39°-13°=128°

As JFE is a straight line ∴ ∠JFG and ∠EFG forms linear pair

⇒ m∠JFG+m∠EFG=180°

⇒ 128°+m∠EFG=180°

⇒ m∠EFG=52°

The measure of ∠EFG is 52°

6 0
3 years ago
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