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Paladinen [302]
2 years ago
13

Given sin0= 3/7, and tan0 <0, what is the value of coso?

Mathematics
1 answer:
spin [16.1K]2 years ago
8 0

sin θ = 3 / 7 = .429

sin^2 θ + cos^2 θ = 1      trig identity

9/49 + cos^2 θ = 1

cos^2 θ = 40 / 49 = .816       cos θ = .903

Check .429^2 + .903^2 = 1

cos .903 = 25.4 deg

In the second quadrant sin = +, cos = -, tan = -

So 25.4 deg in the second quadrant = 180 - 25.4 = 154.6 deg

Check:

sin 154.6 = .428

cos 154.6 = -.903

tan 154.6 = -.475

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The slope of a line is -4 and it’s y-intercept is (0,4) what is the equation of the line that is perpendicular to the first line
Rasek [7]

Original line equation: y=-4x+4

Perpendicular lines have opposite reciprocals, so the slope of the perpendicular line would be 1/4x.

Perpendicular line equation: y=1/4x+b

Plug in the coordinates that the line passes through to solve for b. Plug -8 in for x and 2 in for y.

2=1/4(-8)+b

Multiply.

2=-2+b

Add 2 to both sides.

4=b

Equation of the perpendicular line: y=1/4x+4

Hope this helps! :)

4 0
3 years ago
What is the answer to<br> 56 is 28% of
klio [65]
In mathematics, 'is' would be the equal sign. And 'of' would be the multiplcation sign. The missing number would be any variable.

56 is 28% of a number would be translated algebraically to:

56 = \frac{28}{100} \times x

Solving for x, we get:

56 = \frac{28}{100} \times x

Divide 28 on both sides

2 = \frac{x}{100}

Multiply 100 on both sides

x = 200

56 is 28% of 200
5 0
3 years ago
Read 2 more answers
Given: Parallelogram LMNO; MO ⊥ LN Prove: LMNO is a rhombus. Parallelogram L M N O is shown. Diagonals are drawn from point L to
BaLLatris [955]

Answer: LMNO is a rhombus

Step-by-step explanation: Since LMNO= Mo x LN, Point M=LM x ON, given these circumstances, we know that LMNO is a rhombus from the sides and angles equation.

3 0
3 years ago
Read 2 more answers
Luna ray has 5 orange picks and no green picks you add some green picks so 1 in every 3 picks is orange. How many picks are ther
Black_prince [1.1K]
2 is the answer

I hope this helps!
7 0
2 years ago
World war 1 lasted from August 4th 1914 until November 11th, 1918. During this time an estimated 10 million lives were lost. Abo
Kryger [21]
The question is so dry, mechanical, and devoid of emotion
that it's terrifying.

There is no way to assign a number to "How many people were
dying per day", and I would prefer not even to think about it in
those terms.

-- The period of time from August 4, 1914 until November 11, 1918 is 1,560 days.

-- The "average", or better, the "unit rate" of 10 million events in 1,560 days
is the quotient

                   (10,000,000 events) / (1,560 days) 

               =  6,410.3 events per day

               =     267.1 events per hour

               =       4.45 events per minute.


Reciprocally, this is a unit rate of 

                  13.48 seconds per event,
                  sustained continuously for 4.274 years !

When will we ever learn ! ?
6 0
3 years ago
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