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3 years ago

Use the substitution method to solve each linear system. X - 2y = 7, y = - X + 1

Mathematics

1 answer:

faltersainse [42]3 years ago

7 0

Answer:

x = 3

y = -2

Step-by-step explanation:

x - 2y = 7

y = -x + 1

x - 2y = 7

x - 2 ( -x + 1 ) = 7

3x - 2 = 7

3x = 9

x = 3

y = -x + 1

y = -3 + 1

y = -2

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joseph's cell phone service charges him $0.15 per text. write an equation that represents the cost of c of his cell phone servic

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C=$0.15T say he has 10 texts you replace the t with the ten, so $0.15 • 10= $1.5

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3 years ago

We expect that students who do well on the midterm exam in a course will usually also do well on the final exam. Gary Smith of P

dybincka [34]

Using the model equation, the predicted mean score on the final given a score of 10 points above the class mean in the mid term exam is 50.7

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3 years ago

Find the percent increase from 200 to 240

vlabodo [156]

Answer:

20%

Step-by-step explanation:

The percentage change can be found from ...

% change = ((new value)/(reference value) -1) × 100%

= (240/200 -1) × 100%

= 0.20 × 100%

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240 is an increase of 20% from 200.

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3 years ago

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tatuchka [14]

Answer

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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