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Klio2033 [76]
2 years ago
15

Use trigonometric ratios to find the indicated side

Mathematics
1 answer:
iragen [17]2 years ago
4 0

Answer:

a = 139.1

b = 56.2

Step-by-step explanation:

A. Reference angle = 68°

Opp = a

Hyp = 150

Therefore:

Sin 68 = opp/hyp

Sin 68 = a/150

150*sin 68 = a

a = 139.1 (nearest tenth)

B. Reference angle = 68°

Adj = b

Hyp = 150

Therefore:

Cos 68 = adj/hyp

Cos 68 = b/150

150*cos 68 = b

b = 56.2 (nearest tenth)

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Don't know the exact equation but trying to explain

X is what the stuff that problem saying. Let's say you have books from library and you're returning them late you will pay $30 dollars for each of them

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If a questiong does not tell you to answer in terms of pi do you round
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snow_tiger [21]

the figure moved 7 to the right and 7 down

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2 years ago
The area of a rectangle is 56 cm. the length is 2 less than x and the width is 5 less than twice of x. solve for x
Nostrana [21]

Answer:

x=\frac{9+\sqrt{449}}{4}

Step-by-step explanation:

The area of a rectangle is:

A=l\cdot w

First let's write both the length and the width in terms of x:

l=x-2

<u>How we obtained the answer</u>:

The question states the length is "2 less than x". In this case, "less" is a keyword that represents subtraction. Therefore, "2 less than x" is "x-2".

w=2x-5

<u>How we obtained the answer</u>:

The question states the width is "5 less than twice of x". First, start by figuring out what "twice of x" is. The keyword is "twice" which represents multiplication. Therefore, "twice of x" can be written as "2x". Then, using the same keyword from before ("less"), "5 less than..." represents "2x-5".

Given:

Area = 56 cm

l=x-2

w=2x-5

Find:

x

A=l\cdot w

56=(x-2)(2x-5)

FOIL (Firsts, outsides, insides, lasts):

56=2x^2-4x-5x+10\\56=2x^2-9x+10

Subtract 10 from both sides:

46=2x^2-9x

Subtract 46 from both sides (forms a quadratic equation):

0=2x^2-9x-46

Use the quadratic formula:

{x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}  where 0=ax^2+bx+c

Therefore:

x=\frac{9\pm\sqrt{81-4(2)(-46)}}{2(2)}\\x=\frac{9\pm\sqrt{81+368}}{4}\\\\x=\frac{9\pm\sqrt{449}}{4}

The answer is:

x=\frac{9+\sqrt{449}}{4}

This is because if we used subtraction, we would get a negative value, and we cannot have a negative length/width.

4 0
2 years ago
What is the distance in the standard (x, y) coordinate<br> plane between the points (2,3) and (5,5)?
Naddik [55]
<h3>The distance between the points (2,3) and (5,5) is 3.6 units</h3>

<em><u>Solution:</u></em>

<em><u>Distance between two points is given by:</u></em>

d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

We have to find the distance in the standard (x, y) coordinate  plane between the points (2,3) and (5,5)

From given,

(x_1, y_1) = (2, 3)\\\\(x_2, y_2) = (5, 5)

Substituting the values we get,

d = \sqrt{(5-2)^2 + (5-3)^2}\\\\Simplify\\\\d = \sqrt{3^2 + 2^2}\\\\d = \sqrt{9+4}\\\\d = \sqrt{13}\\\\In\ decimal\ form\\\\d = 3.605 \approx 3.6

Thus distance between the points (2,3) and (5,5) is 3.6 units

3 0
3 years ago
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