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11Alexandr11 [23.1K]
4 years ago
8

The length of a rectangle is increased by 10% and its width is reduced by 20%. Express the area of ​​the new rectangle as a perc

entage of the area of ​​the initial rectangle
Mathematics
1 answer:
Gwar [14]4 years ago
3 0

Answer:

A= (l+0.1l)*(b-0.2b)

Step-by-step explanation:

let the length and width be l and b

length= l +(10/100*l) = l+0.1l

width= b-(20/100*b)= b-0.2b

thus the area would be

A= (l+0.1l)*(b-0.2b)

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In the diagram below of triangle PQR, S is the midpoint of PR and T is the
Nina [5.8K]

Answer:

ST = 4

Step-by-step explanation:

A segment joining the midpoints of 2 sides of a triangle is half the length of the third side.

ST = \frac{1}{2} PQ substitute values

- 32 + 9x = \frac{1}{2} (- 5x + 28) ← multiply both sides by 2 to clear the fraction

- 64 + 18x = - 5x + 28 (add 5x to both sides )

- 64 + 23x = 28 ( add 64 to both sides )

23x = 92 ( divide both sides by 23 )

x = 4

Then

ST = - 32 + 9x = - 32 + 9(4) = - 32 + 36 = 4

5 0
3 years ago
2x + 7y = 3<br> = -4y<br> Y<br> =
yaroslaw [1]
(33/8, -3/4)
( x , y)



4 0
3 years ago
What graph represents an even function
o-na [289]
I don’t c any graphs?
3 0
3 years ago
(PLEASE ANSWER FAST!!!!) Two lines, A and B, are represented by equations given below: Line A: y = x – 2 Line B: y = 3x + 4 Whic
Sophie [7]

Answer:

(-3, -5), because the point satisfies both equations.

6 0
4 years ago
If cosine theta equals negative square root of three over two and pi over two less than theta less than pi, what are the values
Natalka [10]
<h2>Answer:</h2>

The values of sinθ and tanθ are:

\sin \theta=\dfrac{1}{2}

and

\tan \theta=-\dfrac{1}{\sqrt{3}}=-\dfrac{\sqrt{3}}{3}

<h2>Step-by-step explanation:</h2>

It is given that:

\cos \theta=-\dfrac{\sqrt{3}}{2}

where

\dfrac{\pi}{2}

i.e. the angle θ lie in the second quadrant.

Also, we know that in the second quadrant sine and cosecant function is positive.

whereas cosine,secant, tangent and cotangent functions are negative.

Now, we know that:

\cos (\dfrac{5\pi}{6})=-\dfrac{\sqrt{3}}{2}

i.e. here

\theta=\dfrac{5\pi}{6}

Hence, we have:

\sin (\theta)=\sin (\dfrac{5\pi}{6})\\\\i.e.\\\\\sin \theta=\dfrac{1}{2}

Also,

\tan \theta=\dfrac{\sin \theta}{\cos \theta}\\\\i.e.\\\\\tan \theta=\dfrac{\dfrac{1}{2}}{\dfrac{-\sqrt{3}}{2}}\\\\i.e.\\\\\tan \theta=\dfrac{1}{-\sqrt{3}}\\\\i.e.\\\\\tan \theta=-\dfrac{1}{\sqrt{3}}

Now, on rationalizing we have:

\tan \theta=-\dfrac{\sqrt{3}}{3}

7 0
4 years ago
Read 2 more answers
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