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3 years ago

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Emelio wants to solve this equation. x 5 = 35 Help him follow the steps to finish solving for x. 1. Multiplication property of e

quality: StartFraction 5 Over 1 EndFraction times StartFraction x Over 5 EndFraction = 35 times 5 2. Simplify: x =

2 answers:

KiRa [710]3 years ago

6 0

Answer:

175

Step-by-step explanation:

antiseptic1488 [7]3 years ago

3 0

Answer: 175

Step-by-step explanation

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The graph shows the function f(x).<br><br><br><br> Which equation represents f(x)?

VLD [36.1K]

f(x)=-3(x-1)

Answered would be B.

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3 years ago

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How to add and subtract scientific notation

ch4aika [34]

Once the numbers have the same base and exponents, we can add or subtract their coefficients.Here are the steps to adding or subtracting numbers in scientific notation:Determine the number by which to increase the smaller exponent by so it is to the equal larger exponent

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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 $\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)$ .

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3 years ago

Use the reference angles to find the exact value of the expression cot -11 pi/6. Do not use a calculator

laiz [17]

Answer:

$\sqrt{3}$

Step-by-step explanation:

cot(-11pi/6)=cot(-pi-5pi/6)= -cot(5pi/6)=cot(pi/6)= $\sqrt{3}$ ≈1.73

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3 years ago

{y=-x+1<br> {y=4x-14<br> please help me <br> my paper is due today

Temka [501]

$- x + 1 = 4x - 14 \\ - x - 4x = - 14 - 1 \\ - 5x = - 15 \\ x = 3$

Substitute x = 3 in any given equations. I will do in y = - x + 1

$y = - x + 1 \\ y = - 3 + 1 \\ y = - 2$

Therefore the answer is (3,-2)

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3 years ago