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2 years ago

Please help me with this

Mathematics

1 answer:

hoa [83]2 years ago

7 0

Answer: 31 < x < 33

Step-by-step explanation:

It is possible to calculate the value of x using the equation

3x - 9 = 87

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1. What would be the depth of water of 3/4 filled rectangular pool, if pool is 10 m long, 7 m wide and 8 m deep?

Vladimir79 [104]

Step-by-step explanation:

Area if the rectangular pool= 10m×7m×8m

= 560m³

3/4 of 560m³

3/4×560

= 420

Hence , the water occupies 420m³ place

7 0

2 years ago

Jason is playing darts. He hit the bullseye 8 out of his last 30 throws. What is the probability that he does not make a bullsey

Vsevolod [243]

Answer:

Jason lucky

Step-by-step explanation:

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2 years ago

The Diaz family bought 7 bags of cookies. Each bag had 14 cookies. They have since eaten 29 of the cookies. How many cookies do

saveliy_v [14]

Answer:

They have 69 cookies left.

Step-by-step explanation:

7 x 14 = 98

98 - 29 = 69

6 0

3 years ago

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

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

Solve the system: 2x+6y=36 x+4y=20

Assoli18 [71]

Let's solve this problem step-by-step

Let's set:

2x + 6y = 36 -- equation 1

x + 4y = 20 -- equation 2

(equation 2) * 2

2x + 8y = 40 -- equation 3

(equation 3) - (equation 1)

2y = 4

y = 2 -- equation 4

Plug (equation 4)'s value of y into (equation 2)

x + 4(2) = 20

x = 20 - 8

x = 12

Thus x = 12 and y = 2

Let's check, by substituting these values

$2x + 6y = 36\\2(12)+6(2) = 36\\24+12=36\\36=36\\\\x+4y=20\\12+4(2)=20\\12+8=20\\20=20$

Answer: x = 12 and y = 2

Hope that helps!

5 0

2 years ago