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Ostrovityanka [42]
2 years ago
12

Find the sum of 3b/b 2 and 12/b 2 and express it in simplest form

Mathematics
1 answer:
alexandr1967 [171]2 years ago
8 0
\frac{3b}{b^{2}} + \frac{12}{b^{2}} = \frac{3b + 12}{b^{2}} = \frac{3(b) + 3(4)}{b^{2}} = \frac{3(b + 4)}{b^{2}}
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Complete the sequence: 1, 1, 2, 3, 5, ____, ____, ____.
Elza [17]

Answer:

8, 13, 21

Step-by-step explanation:

This is a part of the Fibonacci sequence. it is where you add the two previous numbers together to get the third.

<u><em>Ex. </em></u>

<u><em>1+1 = 2</em></u>

<u><em>1+2 = 3</em></u>

7 0
3 years ago
What is the answer to this question
Ratling [72]
Hello!

As you cans see, the angle is a right angle, which measures 90°. One of the angles measure 35°

90 - 35 = 55

That means 5x = 55°

Divide both sides by 5

\frac{5x}{5} =  \frac{55}{5}

\framebox{x = 11 \°}
7 0
3 years ago
Read 2 more answers
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
.Mariam went to the store to buy some apples. The price per pound of the apples is $6 per pound and she has a coupon for $2.25 o
8_murik_8 [283]

Answer:

$15.75

$6p-$2.25

4 0
3 years ago
Evaluate 3x2 - 1 when x = 2.
Shalnov [3]

Answer:C or 11

Step-by-step explanation:

3x²-1

First you insert x... which is 2

3(2)²-1

then you square the 2. (multiply it by itself)

3(4)-1

Then you multiply 3 by 4 to get 12

12-1

And then you subtract for your answer

11

4 0
3 years ago
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