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3 years ago

Which situation is best? having

Mathematics

2 answers:

uranmaximum [27]3 years ago

8 0

Answer:

Step-by-step explanation:

beks73 [17]3 years ago

4 0

Answer:

whats the queshton

Step-by-step explanation:

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Please give me the correct answer

Mashutka [201]

Answer:

1st one

Step-by-step explanation:

Because It is just like a right triangle.

Also, look at how the bird is formed, it looks like a right triangle.

Also, I know it's a right triangle because of the length given.

Hope this helped!

4 0

3 years ago

Read 2 more answers

What is 300/250 in simplest form

Semenov [28]

300/250 can be divide by 50 which makes it 6/5

7 0

3 years ago

Read 2 more answers

For 0 ≤ ϴ < 2π, how many solutions are there to tan(StartFraction theta Over 2 EndFraction) = sin(ϴ)? Note: Do not include va

Black_prince [1.1K]

Answer:

3 solutions:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

Step-by-step explanation:

So, first of all, we need to figure the angles that cannot be included in our answers out. The only function in the equation that isn't defined for some angles is $tan(\frac{\theta}{2})$ so let's focus on that part of the equation first.

We know that:

$tan(\frac{\theta}{2})=\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

therefore:

$cos(\frac{\theta}{2})\neq0$

so we need to find the angles that will make the cos function equal to zero. So we get:

$cos(\frac{\theta}{2})=0$

$\frac{\theta}{2}=cos^{-1}(0)$

$\frac{\theta}{2}=\frac{\pi}{2}+\pi n$

$\theta=\pi+2\pi n$

we can now start plugging values in for n:

$\theta=\pi+2\pi (0)=\pi$

if we plugged any value greater than 0, we would end up with an angle that is greater than $2\pi$ so, that's the only angle we cannot include in our answer set, so:

$\theta\neq \pi$

having said this, we can now start solving the equation:

$tan(\frac{\theta}{2})=sin(\theta)$

we can start solving this equation by using the half angle formula, such a formula tells us the following:

$tan(\frac{\theta}{2})=\frac{1-cos(\theta)}{sin(\theta)}$

so we can substitute it into our equation:

$\frac{1-cos(\theta)}{sin(\theta)}=sin(\theta)$

we can now multiply both sides of the equation by $sin(\theta)$

so we get:

$1-cos(\theta)=sin^{2}(\theta)$

we can use the pythagorean identity to rewrite $sin^{2}(\theta)$ in terms of cos:

$sin^{2}(\theta)=1-cos^{2}(\theta)$

so we get:

$1-cos(\theta)=1-cos^{2}(\theta)$

we can subtract a 1 from both sides of the equation so we end up with:

$-cos(\theta)=-cos^{2}(\theta)$

and we can now add $cos^{2}(\theta)$

to both sides of the equation so we get:

$cos^{2}(\theta)-cos(\theta)=0$

and we can solve this equation by factoring. We can factor $cos(\theta)$ to get:

$cos(\theta)(cos(\theta)-1)=0$

and we can use the zero product property to solve this, so we get two equations:

Equation 1:

$cos(\theta)=0$

$\theta=cos^{-1}(0)$

$\theta={\frac{\pi}{2}, \frac{3\pi}{2}}$

Equation 2:

$cos(\theta)-1=0$

we add a 1 to both sides of the equation so we get:

$cos(\theta)=1$

$\theta=cos^{-1}(1)$

$\theta=0$

so we end up with three answers to this equation:

$\theta={0, \frac{\pi}{2}, \frac{3\pi}{2}}$

7 0

2 years ago

Some helppp me plzzz

Eddi Din [679]

Answer:

-2, 4

-1, -2

0, 0

1, 2

2, 4

5 0

3 years ago

Read 2 more answers

3. Determine which of the following pairs is an adjacent side for quadrilateral MOLE below.

yKpoI14uk [10]

Pairs which is Adjacent side for quadrilateral MOLE is given below.

Step-by-step explanation:

Given:

Quadrilateral MOLE

Pair of adjacent sides of the quadrilateral.

Adjacent sides have one vertex common.

Option A: MO and LE

These sides does not have common vertex.

MO and LE are opposite sides in the quadrilateral MOLE.

It is not true.

Option B: EO and ME

In the quadrilateral, ME is not a side.

So it is not true.

Option C: LE and OL

In the quadrilateral, OL is not a side.

So it is not true.

Option D: ML and LE

These sides have common vertex L.

Therefore ML and LE are pair of adjacent sides.

It it true.

Hence ML and LE is a pair of adjacent side for quadrilateral MOLE.

7 0

3 years ago