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masha68 [24]
3 years ago
13

A rectangle is three centimeters.longer than it is wide. if its length were to be decreased by two centimeters, its area would d

ecrease by thirty square centimeters.
A) what are the dimensions?
B) what is it’s area?
Show your work using an equation
Mathematics
1 answer:
Scorpion4ik [409]3 years ago
7 0
Given/equations formed from text:
I: l=w+3
II: A=l*w
III: new area B=(l-2)*w
IV: B=A-30

substitute l in II with I to remove l:
II': A=l*w=
(w+3)*w=
w^2+3w

substitute l in III with I to remove l:
III': B=(l-2)*w=
(w+3-2)*w=
(w+1)*w=
w^2+w

substitute A and B from II' and III' into IV:
B=A-30
w^2+w=w^2+3w-30
30=2w
w=15

insert w=15 into I:
l=15+3=18
->
A) dimensions of full size rectangle are width 15 and length 18
dimensions of reduced size rectangle are width 15 and length 16

B) full size: 15*18=10*18+5*18=180+90=270cm^2
reduced size: 15*16=10*16+5*16=160+80=240cm^2


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Consider a triangle ABC like the one below. Suppose that a = 31, b = 23, and c = 20. (The figure is not drawn to scale.) Solve t
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The solution to the triangle is A = 92.0, B = 47.9 and C = 40.1

<h3>How to solve the triangle?</h3>

The figure is not given;

However, the question can still be solved without it

The given parameters are:

a = 31, b = 23, and c = 20

Calculate angle A using the following law of cosine

a² = b² + c² - 2bc * cos(A)

So, we have:

31² = 23² + 20² - 2 * 23 * 20 * cos(A)

Evaluate

961 = 929 - 920 * cos(A)

Subtract 929 from both sides

32 =- 920 * cos(A)

Divide both sides by -920

cos(A) = -0.0348

Take the arc cos of both sides

A = 92.0

Calculate angle B using the following law of sine

a/sin(A) = b/sin(B)

So, we have:

31/sin(92) = 23/sin(B)

This gives

31.0189 = 23/sin(B)

Rewrite as:

sin(B) =23/31.0189

Evaluate

sin(B) =0.7415

Take arc sin of both sides

B = 47.9

Calculate angle C using:

C = 180 - 92.0 - 47.9

Evaluate

C = 40.1

Hence, the solution to the triangle is A = 92.0, B = 47.9 and C = 40.1

Read more about triangles at:

brainly.com/question/2217700

#SPJ1

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