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Kisachek [45]
3 years ago
11

Complete the following statement.

Mathematics
2 answers:
allochka39001 [22]3 years ago
7 0

Answer:

Answers?

Step-by-step explanation:

Hopefully this is the correct answer i'm sorry if it's wrong.

dangina [55]3 years ago
3 0

Answer:

ukkkk

Step-by-step explanation:

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Please assist on two questions, is offering 30+ points!
slava [35]
1. solve for X and y using both equations to find the point they cross in
−x − y = −5
y = x + 1
Plug in y into first equation
-x-(x+1)=-5
-x-x-1=-5
-2x-1=-5
+1 both sides
-2x=-4
÷-2 both sides
x=2 solve for y
y=2+1
y=3

(2,3)

2. you can set them equal to each other so 2x+4=3x+1
3 0
3 years ago
Solve the equation. 36^6x-1 = 6^5x
loris [4]
The answer of the question which you asked above is
36^6x-6^5x=1
7 0
3 years ago
Which of the following is equivalent to 8x2+20x+
kkurt [141]
<h2>answer </h2><h2>36</h2>

<h2>i HOPE IT'S HELP </h2>

7 0
3 years ago
-4 times a number plus 29
kipiarov [429]
-4x+29? I think that's what you're asking.
8 0
3 years ago
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
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