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zaharov [31]
2 years ago
13

Can someone help me please?

Mathematics
1 answer:
Alik [6]2 years ago
5 0

Answer:

C.

Step-by-step explanation:

The reds are counting up in whole numbers by one integer.

Hope this helps!

If not, I am sorry.

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Anna35 [415]
2 ounces per month. 320 divided by 16 (that’s how many ounces are in a pound) divided by 12 (that’s how many months are in a year)
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3 years ago
It has a diameter of 2 inches what is the radius
ValentinkaMS [17]
Radius is the half of diameter.
That means the radius is 2inch/2=1inch.

Good luck mate!
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You are building a ramp that must cover a horizontal distance of exactly 17 feet. The angle of the ramp from the ground is 24°.
tiny-mole [99]

Answer:

Length of the ramp = 18.6 feet (Approx.)

Step-by-step explanation:

Given:

Horizontal distance (base distance) = 17 feet

Angle of elevation = 24°

Find:

Length of the ramp

Computation:

Cosθ = Length of base / Length of hypotenuse

Cos 24 = Horizontal distance (base distance) / Length of the ramp

Cos 24 = 17 / Length of the ramp

0.9135 = 17 / Length of the ramp

Length of the ramp = 17 / 0.9135

Length of the ramp = 18.6 feet (Approx.)

6 0
3 years ago
Question (c)! How do I know that t^5-10t^3+5t=0?<br> Thanks!
astra-53 [7]
(a) By DeMoivre's theorem, we have

(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta

On the LHS, expanding yields

\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta

Matching up real and imaginary parts, we have for (i) and (ii),


\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta
\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta

(b) By the definition of the tangent function,

\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}
=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}

=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}
=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}


(c) Setting \theta=\dfrac\pi5, we have t=\tan\dfrac\pi5 and \tan5\left(\dfrac\pi5\right)=\tan\pi=0. So

0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}

At the given value of t, the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.


0=t^5-10t^3+5t\implies0=t^4-10t^2+5

Remember, this is saying that

0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5

If we replace \tan^2\dfrac\pi5 with a variable x, then the above means \tan^2\dfrac\pi5 is a root to the quadratic equation,

x^2-10x+5=0

Also, if \theta=\dfrac{2\pi}5, then t=\tan\dfrac{2\pi}5 and \tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0. So by a similar argument as above, we deduce that \tan^2\dfrac{2\pi}5 is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)

Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5

But \tanx>0 for all 0, as is the case for x=\dfrac\pi5 and x=\dfrac{2\pi}5, so we choose the positive root.
3 0
3 years ago
Finding Side Lengths in a Right Triangle<br> What is the value of s?<br> units
zavuch27 [327]

Answer:

hypotenuse=root of( base square + height square)

Step-by-step explanation:

by using pythogorus theorem we can find the side of any right triangle square

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