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Aneli [31]
3 years ago
11

Solve each triangle for the missing values shown. Show all work necessary. Use exact values

Mathematics
1 answer:
Gelneren [198K]3 years ago
7 0

3)

Cos 60°= \frac{1}{2} = \frac{Adjacent} {Hypotenuse} = \frac{y} {10} \\\\ 2y=10 \\\\ y=5

Now, use Pythagoras theorem to find x

x^2+y^2=10^2 \\\\ \to x^2+5^2=10^2 \\\\ \to x^2+25=100 \\\\ \to x^2 = 100-25 \\\\ \to x^2= 75 \\\\ \to x= \sqrt{75} \\\\ \bf 5 \sqrt{3}

Similarly,

4)

Sin 45°= \frac{1}{\sqrt{2}} = \frac{Opposite} {Hypotenuse} =\frac{2 \sqrt{2}} {a} \\\\ a= 2 \sqrt{2} \times \sqrt{2} \\\\ a=4

Now, use Pythagoras theorem to find b

b^2+(2 \sqrt{2}) ^2=a^2 \\\\ \to b^2+8=4^2 \\\\ \to b^2=16-8  \\\\ \to b^2 = 8\\\\ \to b= \sqrt{8} \\\\ \bf b=2 \sqrt{2}

You can solve all of them in a similar way.

: )

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6 0
3 years ago
HELP NEEDED !<br> For 11 and 12 if you don’t know please don’t answer this is a benchmark
dalvyx [7]

Answer:

Step-by-step explanation:

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3 0
4 years ago
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HELP!!! I WILL GIVE BRAINLIEST!!! 10 PTS!!!
tatiyna

Answer:

x= 55 degrees

w = 5

Step-by-step explanation:

Triangle PQR contains two triangles, QPS and PSR.

Looking at triangle QPS, it is an isosceles triangle. This is shown by the two short lines that cuts line QP and line QS.

In an isosceles triangle, two sides and two angles are equal. The two equal angles are the base angles.

Therefore,

angle P and angle S are equal. Also

Line QP and line PS are equal. This means

6w - 10 = w

6w-w = 10

5w = 10

w = 10/2 = 5

Angle Q + 100 degrees = 180 ( sum of angles on a straight line)

Angle Q = 180 - 100 = 80 degrees

Angle Q = Angle S(base angles of an isosceles triangle)

Angle PSR = 180 - 80 = 100 degrees( sum of angles on a straight line is 180)

Angle PSR + x + 25 = 180

x = 180 - 25 - 100 = 55 degrees

6 0
3 years ago
HELP NEEDED ASAP. "ONLY ANSWER IF YOU KNOW THE ANSWER"
valkas [14]

Answer:

x =-5±  2i sqrt(14)

Step-by-step explanation:

x^2 + 10x= -81

Take the coefficient of x and divide by 2

10/2 = 5

Square it

5^2 =25

add to each side

x^2 + 10x+25 = -81+25

( x+5) ^2 = -56

Take the square root of each side

sqrt(( x+5) ^2) =± sqrt(-56)

x+5 = ± sqrt(-1) sqrt(4*14)

x+5 =± i sqrt(4)sqrt(14)

x+5 =± i 2 sqrt(14)

Subtract 5 from each side

x+5-5 =-5± i 2 sqrt(14)

x =-5±  2i sqrt(14)

4 0
3 years ago
The UCL and LCL for an x-bar chart are 25 and 15 respectively and the centerline is 20. The process range is considered to be in
son4ous [18]

Answer:

nothing

Step-by-step explanation:

you do not have to do anything given these seven sample means. This is because the sample means are all within the upper and the lower control limits. the plots are randomized and they do not form any patterns. This process is still in control so you do not really have to do anything. This answers the question.

3 0
3 years ago
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