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Novosadov [1.4K]
3 years ago
6

Solve the equation. n+34=14

Mathematics
2 answers:
Lunna [17]3 years ago
8 0

Answer:

n = -20

Step-by-step explanation:

n + 34 = 14

--------------------

n + 34 - 34 = 14 - 34

n = -20

wlad13 [49]3 years ago
4 0

Answer:

n = -20

Step-by-step explanation:

Subtract 34 on both sides of the equation:

n = 14 - 34

n = -20

Hope this helps!

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Step-by-step explanation:

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3 years ago
Let A=(-6,7), B=(-4,11), and C=(-2.10) be three points in the coordinate plane (A) Verify that the three points form a right tri
faust18 [17]

Answer:

  (A) see below

  (B) the right angle is at vertex B(-4,11)

  (1) 3x -4y = -46

  (E) the midpoint is (-4, 8.5)

Step-by-step explanation:

(A) The slope of line AB is Δy/Δx = (11-7)/(-4-(-6)) = 4/2 = 2. The slope of line BC is (10-11)/(-2-(-4)) = -1/2. The slopes of AB and BC have a product of 2(-1/2) = -1, so are the slopes of perpendicular lines. The points are distinct, and lines joining two pairs of them are at right angles, so the points form a right triangle.

(B) Point B(-4,11) is the point of intersection of perpendicular segments AB and BC, so is the location of the right angle.

(1) The slope of the hypotenuse AC is ...

  Δy/Δx = (10-7)/(-2-(-6)) = 3/4

In point-slope form, the equation for the line through point A with this slope is ...

  y -7 = 3/4(x +6) . . . . point-slope form of the equation of the hypotenuse

  4y -28 = 3x +18 . . . . multiply by 4

  -46 = 3x -4y . . . . .  . subtract 18+4y

  3x -4y = -46 . . . . . . . standard form equation of the hypotenuse

(E) The midpoint of the hypotenuse is the average of the endpoint coordinates:

  M = (A + C)/2 = (-6-2, 7+10)/2 = (-4, 8.5)

The midpoint of the hypotenuse is M(-4, 8.5).

8 0
4 years ago
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Answer:

see explanation

Step-by-step explanation:

We require to find the third side of the triangle.

Using Pythagoras' identity

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let x represent the third side, then

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x² + 225 = 289 ( subtract 225 from both sides )

x² = 64 ( take the square root of both sides )

x = \sqrt{64} = 8

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tanΘ = \frac{opposite}{adjacent} = \frac{15}{8}

cosΘ = \frac{adjacent}{hypotenuse} = \frac{8}{17}

sinΘ = \frac{opposite}{hypotenuse} = \frac{15}{17}

cscΘ = \frac{1}{sin0} = \frac{1}{\frac{15}{17} } = \frac{17}{15}

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