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zhannawk [14.2K]
2 years ago
9

(-2)^5 !!!!!!!!!! help

Mathematics
2 answers:
aalyn [17]2 years ago
8 0

Answer:

(-2)^5 = -32

Step-by-step explanation:

(-2)^5 = -2 • -2 • -2 • -2 • -2 = -32

Natali5045456 [20]2 years ago
7 0

Answer:

-32

Step-by-step explanation:

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Step-by-step explanation:

-1+8=7

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Answer: 72 feet

Step-by-step explanation:

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Find the unit rate.<br><br> 6 2/5 revolutions in 2 2/3 seconds
topjm [15]
1) Convert mixed fractions into fractions.

6 2/5 = ((6*5) + 2) / 5 = 32 / 5
2 2/3 = ((2*3) + 2) / 3 =  8 / 3

2) 32 / 5 ÷ 8 / 3 ; where 32/5 is the 1st fraction and 8/3 is the 2nd fraction
a) Get the reciprocal of the 2nd fraction:
   From 8/3 to 3/8
b) Multiply 1st fraction to the reciprocal of the 2nd fraction
  32 / 5 * 3 / 8 = (32*3) / (5*8) = 96 / 40
c) Simplify the fraction.
96 / 40 divide by 4 will become 24/10.
24 / 10 divide by 2 will become 12 / 5. The simplified fraction of 96/40.

 The unit rate is 12 / 5 = 2 2/5 revolutions per second.
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ANSWER ASAP The area of a rectangle depends on its length and width. For the area to remain constant, the length varies inversel
Free_Kalibri [48]
Area of the rectangle is equal to length times width

In function notation, the Area of the rectangle function is a function in variables l (length) and w(width) as follows:

A(l, w)= l*w

for A=475 we have,

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Answer: l= \frac{475}{w} , w= \frac{475}{l} 


4 0
3 years ago
y=c1e^x+c2e^−x is a two-parameter family of solutions of the second order differential equation y′′−y=0. Find a solution of the
vagabundo [1.1K]

The general form of a solution of the differential equation is already provided for us:

y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},

where c_1, c_2 \in \mathbb{R}. We now want to find a solution y such that y(-1)=3 and y'(-1)=-3. Therefore, all we need to do is find the constants c_1 and c_2 that satisfy the initial conditions. For the first condition, we have:y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.

For the second condition, we need to find the derivative y' first. In this case, we have:

y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.

Therefore:

y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.

This means that we must solve the following system of equations:

\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.

If we add the equations above, we get:

\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e}  \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.

If we now substitute c_1 = 0 into either of the equations in the system, we get:

c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}

This means that the solution obeying the initial conditions is:

\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.

Indeed, we can see that:

y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3

y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,

which do correspond to the desired initial conditions.

3 0
3 years ago
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