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3 years ago

Help me please

Mathematics

1 answer:

kari74 [83]3 years ago

6 0

Answer:

$\frac{ \cos(80) }{ \sin(10) } + \frac{ \sin(20) }{ \cos(70) } = 2 \\ \frac{ \cos(90 - 10) }{ \sin(10) } + \frac{ \sin(20) }{ \cos(9 0 - 20) } = 2 \\ \frac{ \sin(10) }{ \sin(10) } + \frac{ \sin(20) }{ \sin(20) } = 2 \\ 1 + 1 = 2 \\ 2 = 2 \\ \\ \frac{ \cot(40) }{ \tan(50) } + \frac{ \cos(65) }{ \sin(115) } = 2 \\ \frac{ \cot(90 - 50) }{ \tan(50) } + \frac{ \cos(65) }{ \sin(90 + 65) } = 2 \\ \frac{ \tan(50) }{ \tan(50) } + \frac{ \cos(65) }{ \cos(65) } = 2 \\ 1 + 1 \\ = 2$

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user100 [1]

Using Pythagorean theorem, we can justify our answer.
if c^2 = a^2 + b^2 then triangle is right one,
if c^2 > a^2 + b^2 then triangle is obtuse and
if c^2 < a^2 + b^2 then triangle is acute triangle.

Here a=8, b=11 and c=16
Put these values in the equation
16^2 = 8 ^2 + 11^2
256 = 64 +121
256> 185
So here c^2 > a^2 + b^2 Which means triangle is obtuse triangle.

Answer: Obtuse Triangle

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3 years ago

1. Write a mixed number that is equivalent to 16

Bumek [7]

Answer:

15 4/4

Step-by-step explanation:

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3 years ago

I need help on number 18

zloy xaker [14]

Answer:

Volume = 35185.8 $km^{3}$

Step-by-step explanation:

Recall the formula for the volume of a cone: $Volume = \pi R^{2} \frac{h}{3}$

Where R is the radius of the cone's circular base (in your case half of 80 km,(that is 40 km)

And h stands for the height of the cone (in your case 21 km)

We replace such in the formula and calculate its final value:

$Volume = \pi 40^{2} \frac{21}{3} = \pi 1600 \frac{21}{3} = \pi 1600 (7) = 35185.83772...$

We round it to the number of decimals your problem request (let's say to one decimal) and also include the cubic units:

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3 years ago

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Alex787 [66]

Answer:475

Step-by-step explanation:

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118.75x10=475

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3 years ago

How could you change the equation from Part B so it has infinitely many solutions? What would infinitely many solutions mean in

Andrews [41]

Answer:

We do not have the equation for the part B, so we can not aswer it correctly.

But i will give a general answer.

We could have infinite answers always when we have more variables than linear independent equations:

This is, if we have one variable, x, we can have infinite solutions if we have no equations (or equations with no restrictions for our variable)

So if we have an equation like:

x*4 = √16*x

you can see that both sides of the equation are exactly the same, so this equation actually does not have any value, and x could take infinite different values and the equation will remain true.

If we have two variables, x and y, we will have infinite solutions if we have only one equation:

y = a*x + b

We have infinite pairs (x, y)

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3 years ago