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enyata [817]
3 years ago
10

Which ordered pair is a solution of the equation? -3x – y = 6 Choose 1 answer:

Mathematics
1 answer:
Aleks [24]3 years ago
5 0

Answer:

Step-by-step explanation:

I can not see your answers to choose but with this equation for 2 variables, you can replace the values of x and y and see if that is right or not.

For example, to know (-2,5) is a solution, you replace x= -2 and y= 5

into this equation: -3x -y = 6

and you have: -3*(-2) -5 =6

or 6-5 =6, and you find that is impossible, so (-2.5) is not a solution.

Hope you understand it

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X<br>+ 3 + 5x =<br>I dont know<br>understand<br>this help me​
spin [16.1K]

Answer:

6x + 3

Step-by-step explanation:

Just add the X's (5x + x)=6x, since there is only one whole number without any variables you dont have to do anything to it. So 6x + 3 is your answer.

4 0
3 years ago
Read 2 more answers
Fh
scZoUnD [109]

Answer:

ok so just get the v12{{9-))0} and =14

Step-by-step explanation:

4 0
2 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
What is 16 to 29 and 15 in at ratio
Semmy [17]
16:29:15 I think that would be it but, I'm really sorry if that's not right
7 0
3 years ago
Use the Pythagorean theorem to find the missing area
fgiga [73]

Answer:

28 cm^2

Step-by-step explanation:

a^+21^2=36^2

a^2+441=1225

a=28

5 0
3 years ago
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