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grin007 [14]
3 years ago
8

How do I create a radical equation with an extraneous solution quickly? How do I go the same for a radical equation with a non-e

xtraneous solution?
Mathematics
1 answer:
Step2247 [10]3 years ago
5 0

Answer:

why r u wafflin

Step-by-step explanation:

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Jones of Boston borrowed $40,000, 90 day, 10% note. After 60 days Jones made an initial payment of $6,000. Assuming the U.S. Rul
nadezda [96]
check a website called math-way without the dash it might help
6 0
3 years ago
Select the end behavior of the exponential function shown in the graph
Ber [7]

Answer:

The x graph

Step-by-step explanation:

The x graph

8 0
3 years ago
Read 2 more answers
find the area and perimeter using the lengthy and with step by step I will give 40 points please answer right number 6,7,9,10,11
N76 [4]

Answer:

6. 16 meters^2, 16 meters

7. 100 milimieters^2, 40 milimeters

9. 56 centimeters^2, 30 centimeters

10. 49 inches^2, 28 inches

11. 9 meters^2, 12 meters

12. 72 yards^2, 36 yards

Step-by-step explanation:

s is for a square

A=s^2

P=4s

l+w is for a rectangle

A=lw

P=2l+2w

6. A=4^2

A= 16 meters^2

P=4*4

P= 16 meters

7. A=10^2

A=100 milimieters^2

P=4*10

P= 40 milimeters

9. A=8*7

A= 56 centimeters^2

P=2(8)+2(7)

P=30 centimeters

10. A=7^2

A=49 inches^2

P=4*7

P=28 inches

11. A= 3^2

A=9 meters^2

P=4*3

P= 12 meters

12.

A=12*6

A=72 yards^2

P= 2(12)+2(6)

P= 36 yards

Let me know if you have any questions. Also, don't forget units!

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
How many points do you get if you answer this question ? ​
GrogVix [38]

Answer:

what question ?

Step-by-step explanation:

i did not get it what question?

7 0
3 years ago
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