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2 years ago

PLZ HELP ME WITH THIS QUESTION!

Mathematics

1 answer:

PtichkaEL [24]2 years ago

7 0

Answer:

The correct answer is A, 3,600 yards.

Step-by-step explanation:

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The scatter plot below shows a linear association. Which is the best linear model for the data?

Ivanshal [37]

Answer:

ye its y=5x-10

Step-by-step explanation:

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3 years ago

(6.25•10^9)(9.8•10^16)

AVprozaik [17]

Answer:

This is your answer.. you add exponents

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3 years ago

What is the equation of the line that passes through the points (-1 , 3/2) and (1/2 , 1/2 )

Simora [160]

Answer:

y = -2/3x + 5/6 (slope intercept) (y = ax + b)

4x + 6y = 5 (standard form) (Ax + By = C)

y - 3/2 = -2/3(x + 1) (point slope) (y-y1 = m(x-x1)) (m = y2 - y1 / x2 - x1)

______________________________________

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3 years ago

Find the probability of landing on an even number when spinning the spinner below.

Mariulka [41]

Answer:

There are 8 numbers.

4 of the numbers are even ( 2, 4, 6, 8)

The probability is the number of even numbers over the total numbers:

4/8 which reduces to 1/2.

The answer is 1/2

3 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:

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

7 0

3 years ago