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3 years ago

Which of the lines graphed isn’t he diagram represents the equation x - 3y = -3?

Mathematics

1 answer:

stira [4]3 years ago

4 0

Answer: line d
step by step explanation:

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(Please show all work)

Lunna [17]

The answers here are

2) x= 2, 7

3) x= -5

7 0

4 years ago

The product of two consecutive integers is 342. What quadratic equation can be used to find x, the grater number?

IceJOKER [234]

X (X - 1) = 342
x² - x = 342
x² - x - 342 = 0

X = 19

5 0

4 years ago

Mindy and Troy combined ate 999 pieces of the wedding cake. Mindy ate 333 pieces of cake and Troy had 1/4 of the total cake. W

lubasha [3.4K]

Answer:

Step-by-step explanation:

333+p/4=999

p/4=666

p=2664

6 0

3 years ago

Read 2 more answers

Please answer this question

Gelneren [198K]

A=2(LW+LH+WH)

A=2((7/8)(1/3)+(7/8)(2/5)+(1/3)(2/5))

A=2(7/24+14/40+2/15)

A=14/24+28/40+4/15

A=7/14+7/10+4/15 210

A=(105+147+56)/210

A=308/210

A=(210+98)/210

A=1 98/210

A=1 7/15

8 0

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:

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

7 0

3 years ago