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3 years ago

Can you help me with this??

Mathematics

1 answer:

svetoff [14.1K]3 years ago

7 0

9514 1404 393

Answer:

relative maxima: x=-3, +1
relative minima: x=-1, +5

Step-by-step explanation:

The relative maxima are the peaks, the highest points. There are two of them. One is to the left of the y-axis at about x=-3; the other is to the right of the y-axis at about x=1.

The relative minima are the valleys, the lowest points. There are two of those, also. One is to the left of the y-axis at about x=-1; the other is to the right of the y-axis at about x=5.

relative maxima: x=-3, +1
relative minima: x=-1, +5

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This is confusing i need help ;-;

katovenus [111]

Answer:

B I think

Step-by-step explanation:

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3 years ago

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Brainly annie has 2 bags of apples and 5 single apples eva has 1 bag of apples and 11 single apples they have the same amount of

bekas [8.4K]

Answer:

34 apples

Step-by-step explanation:

To answer: try setting up each of the equations for Annie and Eva.

Annie: 2(bag) + 5

Eva: 1(bag) + 11

Those two equations equal each other. So set that up:

2(bag) + 5 = 1(bag) + 11

Now solve for the size of the bag.

2(bag) - 1(bag) = 11 - 5

1(bag) = 6 apples

Now substitute 6 back into either equation to determine number of apples.

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They each have 17, so there are 34 apples in total.

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3 years ago

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kumpel [21]

Lawrence's is the correct answer. he put 3 plates on each box in both situations

8 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

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4 years ago

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Anon25 [30]

Answer: y = 108

Step-by-step explanation:

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