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Naily [24]
3 years ago
14

Simplify the expression. -2(7-12)

Mathematics
2 answers:
Sladkaya [172]3 years ago
7 0
-2(7-12)=
-2(-5)=
10 = Answer
mylen [45]3 years ago
6 0
Answer: 10
7 minus 12 is negative 5. Negative 2 times negative 5 is 10. 
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Show work <br> Need fast!!! <br><br> 4(1/3+ 1/3+ 1/3)
aalyn [17]

Answer:

4

Step-by-step explanation:

calculate the sum 4*1 any expression is multiplied by 1 remains the same solution is 4

7 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
the number of chirps made by a cricket varies directly as the temperature. If at 12 degrees a cricket chirps 30 times per minute
GenaCL600 [577]

Answer:

50 i think?

Step-by-step explanation:

30 divided by 12 is 2.5, 2.5 x 20 equals 50

4 0
3 years ago
Peter has 3200 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed a
salantis [7]

Answer:

A = 640000\,yd^{2}

Step-by-step explanation:

Expression for the rectangular area and perimeter are, respectively:

A (x,y) = x\cdot y

3200\,yd = 2\cdot (x+y)

After some algebraic manipulation, area expression can be reduce to an one-variable form:

y = 1600 -x

A (x) = x\cdot (1600-x)

The first derivative of the previous equation is:

\frac{dA}{dx}= 1600-2\cdot x

Let the expression be equalized to zero:

1600-2\cdot x=0

x = 800

The second derivative is:

\frac{d^{2}A}{dx^{2}} = -2

According to the Second Derivative Test, the critical value found in previous steps is a maximum. Then:

y = 800

The maximum area is:

A = (800\,yd)\cdot (800\,yd)

A = 640000\,yd^{2}

8 0
3 years ago
Read 2 more answers
I need help with a question
kondaur [170]
What question do you need help with
5 0
3 years ago
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