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Naily [24]
3 years ago
5

Given: log2 = a, log7 = b. Find: log56

Mathematics
1 answer:
madam [21]3 years ago
7 0

\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ log_a(xy)\implies log_a(x)+log_a(y) \end{array} ~\hfill \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ log_a\left( x^b \right)\implies b\cdot log_a(x) \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}


\bf log(56)~~ \begin{cases} 56=2\cdot 2\cdot 2\cdot 7\\ \qquad 2^2\cdot 2\cdot 7 \end{cases}\implies log(2^2\cdot 2\cdot 7) \\\\\\ log(2^2)+\stackrel{a}{log(2)}+\stackrel{b}{log(7)}\implies 2\stackrel{a}{log(2)}+a+b\implies 2a+a+b\implies 3a+b

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How to know if a function is periodic without graphing it ?
zhenek [66]
A function f(t) is periodic if there is some constant k such that f(t+k)=f(k) for all t in the domain of f(t). Then k is the "period" of f(t).

Example:

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

\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5

More generally, it can be shown that

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

is periodic with period \mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n).
4 0
3 years ago
Write the number shown by 200+10+7
Natasha_Volkova [10]
The number would be 217
8 0
3 years ago
Read 2 more answers
How do I to problems like "From 45 ft to 92 ft"?
djyliett [7]

Answer:

45tf+47tf=92tf

Step-by-step explanation:

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3 0
3 years ago
I need this asap. Find the area of the following quadrilateral.
Elenna [48]

Answer:56

Step-by-step explanation:

i got it correct on my test

4 0
2 years ago
Instructions in picture
Kipish [7]

Answer:

12)

(9y +7)=(2y +98)( Because vertically opposite angle is always equal)

9y - 2y = 98-7

9y - 2y = 98-7

7y= 91

y =13

<em>ther</em><em>fore</em><em> </em><em>y</em><em> </em><em>=</em><em> </em><em>1</em><em>3</em>

<em>(</em><em>9</em><em>y</em><em> </em><em>+</em><em> </em><em>7</em><em>)</em>

9*13+7

117+7

124

(2y +98)

2*13+98

26+98

124

5 0
3 years ago
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