Common denominators for 14 and 710

Teresa is factoring this polynomial by grouping. Which common factors should be used in the next step of factoring? 10x3 3x2−20x

Oksi-84 [34.3K]

Answer:

−2=−2x^3

Step-by-step explanation:

10x3= 3x2−20x−6=(10x3⋅(3x2))(−20x−6)=−600x6−180x5x2=−600x6−180x5−2x⋅(2x2)=−4x3−2x⋅x2=−2x^3

−2=−2x3

2x^2=−2x^3

−2=−2x^3

8 0

2 years ago

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Find the gain percent in the following: Cost Price Rs 280, (a) Selling Price Rs 290

Nikolay [14]

Answer:

3.57

Step-by-step explanation:

P=SP -CP

=290-280=10

P%= $\frac{P*100}{CP}$

= $\frac{10*100}{280}$

= $\frac{1000}{280}$

=3.57

4 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

Plzzzz help meeeeeeeee 45 points ASAAAPPPPPPP

lubasha [3.4K]

Answer:

240

Step-by-step explanation:

If you draw a straight line, 12 would meet at 240. Hope this is helpful! You were close but I think you mixed up 240 and 300.

6 0

2 years ago

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Explain your answer !! Have a nice day Will give braisnlt

Viktor [21]

Answer:

(3,-4) or x=3 and y= -4

Step-by-step explanation:

I'm going to solve this by substitution

We first need to get a variable by itself in one of the two equations (it doesn't matter which variable and the equation you do the work on doesn't matter either)

I'm going to solve for y in the second equation

-4x-4y=4

add 4y and subtract 4 from both sides to get

-4x-4=4y

Divide by 4 to get

-x-1=y

We can plug this value in for y into the first equation and get

4x+5(-x-1)= -8

Solve for x

4x-5x-5= -8

-x-5= -8

-x= -3

x=3

We can plug this value into one of the first two equations and solve for y

4(3)+5y= -8

12+5y= -8

5y= -20

y= -4

Therefore the solution is (3,-4) or x=3 and y= -4

8 0

2 years ago