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Katen [24]
3 years ago
11

Question 19 I know the answer is 1.5 but how

Mathematics
1 answer:
katrin2010 [14]3 years ago
5 0
Suppose the width of the path is x. 
the dimensions of the extended rectangle are (12+x+x), and (16+x+x) 
(12+2x)(16+2x)=285
4x²+56x+192=285
4x²+56x-93=0
Not easy to factor it, so use the quadratic formula to solve:
x={-56+√[(56²-4*4*(-93)]}/2*4 or x={-56-√[(56²-4*4*(-93)]}/2*4
x=1.5 (ignore the negative solution.)
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fenix001 [56]

<em>Length of side YZ to the nearest tenth is 16 ft</em>

<em></em>

To solve this question we would use the sine rule. The sine rule states that

<u>A/sin a = B/sin b</u>

if we take our angle a to be XZY, and b to be ZXY.

Going by angles in a triangle, ZXY = 180 - 90 - 35 = 55. This means that our angle b is 55º

applying the sine rule, we have

11/sin 35 = B/sin 55

11/0.5736 = B/0.8192, if we cross multiply, we have

B = (11 * 0.8192)/0.5736

B = 9.0112/0.5736

B = 15.7 ft

learn more about sine rule here brainly.com/question/9599008

6 0
2 years ago
At a supermarket salad bar, the price of a salad depends on its weight. Salad costs $.19 per ounce. Write a rule to describe the
Makovka662 [10]

Answer:

For an 8-ounce salad would cost $1.52

Step-by-step explanation:

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So you would multiply the cost with how much you would buy.

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7 0
3 years ago
The ideal width of a safety belt strap for a certain automobile is 5 cm. An actual width can vary by at most 0.35 cm. Write and
notsponge [240]

Answer:

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Let the actual width of belt be x

we are given that the ideal width of safety belt strap is 5 cm

So, Difference between actual and ideal = (x-5)cm

We are also given that An actual width can vary by at most 0.35 cm.

So, The difference between actual and ideal width should be less than or equal to 0.35 cm

So, |x-5|\leq 0.35

We know

x-5\leq 0.35 and x-5\geq -0.35

x\leq 0.35+5 and x\geq -0.35+5

x\leq 5.35 and x\geq 4.65

So, 4.65\leq x\leq 5.35

Hence the range of acceptable widths is 4.65\leq x\leq 5.35

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3 years ago
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8 0
3 years ago
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