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Mkey [24]
2 years ago
9

The points I, J, K and L all lie on the same line segment, in that order, such that the ratio IJ:JK:KL is equal to 2:5:5 IfIL=60

, find KL.
Mathematics
1 answer:
Elanso [62]2 years ago
5 0

The length of KL with the given ratio is 25 units

<h3>How to determine the length of KL?</h3>

The given parameters are:

IJ : JK : KL is equal to 2 : 5 : 5

IL = 60

Rewrite properly as:

IJ : JK : KL = 2 : 5 : 5

This means that the length of KL

KL = KL ratio/Sum of ratios * IL

So, we have

KL = 5/(2 + 5 + 5) * 60

Evaluate the sum

KL = 5/12* 60

Divide 60 by 12

So, we have

KL = 5 * 5

Evaluate the product

KL = 25

Hence, the length of KL with the given ratio is 25 units

Read more about line segments at:

brainly.com/question/14366932

#SPJ1

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If f(x)=\sin x, then we have \sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x, and so \sin x is periodic with period 2\pi.

It gets a bit more complicated for a function like yours. We're looking for k such that

\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5

Expanding on the left, you have

\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2

and

1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5

It follows that the following must be satisfied:

\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}

The first two equations are satisfied whenever k\in\{0,\pm4,\pm8,\ldots\}, or more generally, when k=4n and n\in\mathbb Z (i.e. any multiple of 4).

The second two are satisfied whenever k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}, and more generally when k=\dfrac{10n}7 with n\in\mathbb Z (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when k is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2

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More generally, it can be shown that

f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))

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