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Kazeer [188]
3 years ago
14

If (x-1)/(x)=20 then x=

Mathematics
2 answers:
nadezda [96]3 years ago
6 0
D:x\neq0\\\\\frac{x-1}{x}=20\ \ \ \ |cross\ multiply\\\\20x=x-1\ \ \ \ |subtract\ x\ from\ both\ sides\\\\19x=-1\ \ \ \ \ \ |divide\ both\ sides\ by\ 19\\\\\boxed{x=-\frac{1}{19}}

Tanya [424]3 years ago
6 0
(x-1)/(x)=20

20x=x-1

20x-x=-1

19x=-1

x=-(1/19)
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hammer [34]

The answer to sqrt 32, simplified is 4sqrt2.

4 0
3 years ago
Help me pls! Thnx if u do ✨✌️
daser333 [38]
Neither of them are correct. When you plug in both of the fractions into the equation, you get answers like 2/9 (from plugging in 1/3) and 1/9 (from plugging in 1/6). None of them are equal to 9.
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3 years ago
Mary read a book that had 4200 pages last summer. If it took her 35 days to finish the book. How many pages did she read per day
asambeis [7]
Do 4200/35............
5 0
2 years ago
1. ​Find the area of the triangle. Show as much work as possible for full credit.
Ugo [173]

Answer:

A = 5/12 x^2

Step-by-step explanation:

A = 1/2 bh

The base is 5/6x and the height is x

A = 1/2 (5/6x) *x

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5 0
3 years ago
If
arlik [135]

Given:

In a right angle triangle θ is an acute angle and \tan\theta =\dfrac{3}{5}.

To find:

The value of \cos \theta.

Solution:

In a right angle triangle,

\tan \theta=\dfrac{Perpendicular}{Base}

We have,

\tan\theta =\dfrac{3}{5}

It means the ratio of perpendicular to base is 3:5. Let 3x be the perpendicular and 5x be the base.

By using Pythagoras theorem,

Hypotenuse=\sqrt{Perpendicular^2+base^2}

Hypotenuse=\sqrt{(3x)^2+(5x)^2}

Hypotenuse=\sqrt{9x^2+25x^2}

Hypotenuse=\sqrt{34x^2}

Hypotenuse=x\sqrt{34}

In a right angle triangle,

\cos \theta=\dfrac{Base}{Hypotenuse}

\cos \theta=\dfrac{5x}{x\sqrt{34}}

\cos \theta=\dfrac{5}{\sqrt{34}}

Therefore, the value of \cos \theta is \dfrac{5}{\sqrt{34}}.

8 0
2 years ago
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