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galben [10]
2 years ago
7

Please answer: p - 4 = -9 + p

Mathematics
2 answers:
nevsk [136]2 years ago
8 0

Answer:

\mathrm{No\:Solution}

Step-by-step explanation:

p-4=-9+p

Step 1: Subtract p from both sides

p-4-p=-9+p-p\\\\-4=-9

Since negative four is not equal, and does not equal negative 9, the answer is \mathrm{No\:Solution}, NOT that there is NO SOLUTION.

kirill115 [55]2 years ago
4 0

Answer:

There are no values of  p  that make the equation true.

(No solution)

Step-by-step explanation:

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lesya692 [45]

Answer:

Formula to find the area of a triangle is 1/2(bxh). The area of the right triangle is 28.

Step-by-step explanation:

To solve this problem you would plug in the base and height into the formula and follow the order of operations (P.E.M.D.A.S). So the equation would then read 1/2(5x11.2). Once you solve inside the parenthesis, the equation then reads, 1/2(56). Then you solve what half of 56 is, which is 28.


I hoped this helped you!

6 0
3 years ago
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Geometry math question no Guessing and Please show work thank you
vesna_86 [32]

First let us find the length of JN

JN =JK +KN

JN = 82+105 = 187

JN=187..............(1)

UC= JN -( JH +HU+CN)

We are given :

JH= 22, HU = 96 and CN = 51

plugging all the values we get

UC = 187-( 22+96+51)

UC =187 -169

UC = 18

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8 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

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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
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djverab [1.8K]

Answer:

2.5

Step-by-step explanation:

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Leviafan [203]

Answer:

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Step-by-step explanation:

a = p*w

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8 0
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