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3 years ago

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Sienna has $8 and is saving $3 per week. Jacob has $6 and is saving $4 per week. Which model represents the equation that can de

termine when Sienna will have the same amount of money as Jacob?
8 Negative x tiles and 3 positive 1 tiles = 6 positive x tiles and 4 positive 1 tiles

8 x tiles and 3 positive 1 tiles = 6 x tiles and 4 positive 1 tiles

3 x tiles and 8 positive 1 tiles = 4 x tiles and 6 positive 1 tiles

3 negative x tiles and 8 positive 1 tiles = 4 x tiles and 6 positive 1 tiles

2 answers:

slamgirl [31]3 years ago

8 0

Answer:

b

Step-by-step explanation:

8090 [49]3 years ago

7 0

Answer:

B

Step-by-step explanation:

From the given information:

Sienna has $8 denotes the unit blocks of x tiles. So, she saved $3 per week.

This implies that:

8x + 3(1) = 8x + 3

Also,

If Jacob as well had $6 which implies 6x unit block of tiles while he saved $4;

i.e.

6(x) + 4(1) = 6x + 4

So;

the model that can determine when Sienna will have the same amount as Jacob is:

8x + 3 = 6x + 4

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Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

2 years ago

Help pleaseeeee?????

Artyom0805 [142]

They would be identical so it would be 114

4 0

2 years ago

please help :( please help :( please help :( please help :( please help :( please help :( please help :(

bixtya [17]

Answer:

1. 3*8*10=240

2.6*6*12=432

Step-by-step explanation:

Hope that helps

Use l*w*h

4 0

3 years ago

If someone could help me with this entire back side that would be great :)

SpyIntel [72]

Answer:

A.)x/8=3/4

x=6

B.)2/5=x/40

x=16

C.)1/8=x/12

x=3/2 or 1.5 or 1 1/2

D.)x/10=12/15

x=8

A.)x/2+x/6=7

x=21/2 or 10.5 or 10 1/2

B.)x/9+2x/2=1/3

x=3/10 or 0.3

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3 years ago

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A=pi r^2 which if the following equations can be used to solve for the radius (r)

zhannawk [14.2K]

C. You would first have to divide A by Pi, then take the square root of r, in order to get r by itself.

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2 years ago