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3 years ago

11

I need to know the answers because my friends and I cannot have different answer.

1 answer:

dezoksy [38]3 years ago

7 0

Show work i dont show work all the time

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8. Sheila and her study partner flipped a coin 120 times. They flipped 70 heads and 50 tails. What is the

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2 years ago

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Is negative three a perfect cube?

erik [133]

No, it is not a perfect cube. A perfect cube is a number that is obtained when you cube an integer. For example, 8 (cube of 2), 27 (cube of 3) and 64 (cube of 4). Since -3 cannot be obtained by cubing an integer, it is not a perfect cube.

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2 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

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Answer:

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Step-by-step explanation:

$use \: sine \: rule \\ \frac{x}{ \sin(31) } = \frac{42}{ \sin(59) } \: \: taking \: angle \: at \: c \: as \: 90 \\ x = \frac{ 42 \sin(31)}{ \sin(59)} \\ x = 25.236146$

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1 year ago

I'm supposed to write out an algebraic equation, and then solve the equation. If you can help me, with work shown, you'll get vo

Arada [10]

Answer:

simple just follow me and mark it brainliest

Step-by-step explanation:

LET THE THREE CONSECUTIVE EVEN INTEGERS BE

(x+2) ,(x+4) ,(x+6)

then

according to the question

4(x+2) – (x+6) = 2(x+4) + 6

4x + 8 – x – 6= 2x + 8+6

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2 years ago

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Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago