Use a number line to find the value of the expression -4+4-9 what property can you use to make your calculation easier explain

Simplify 14 +{ -2 + 3 [ 1 +3 ( -6 -2 ) ] }

gregori [183]

ANSWER: -57
14+{−2+3[1+3(−6−2)]}
= 14+{−2+3[1+3(-8)]}
= 14+{−2+3[1+(-24)]}
= 14+{−2+3[1-24]}
= 14+{−2+3[-23]}
= 14+{−2+[-69]}
= 14+{−2-69}
= 14+(-71)
= 14 - 71
= -57

5 0

2 years ago

Write the equation described in each case.

Lapatulllka [165]

Answer:

The Answer is: y = -1/4x + 6

Step-by-step explanation:

Given point: (4, 5)

Given equation: y = -1/4x - 11

Slope: m = -1/4, a parallel line will have the same slope.

Start with the point slope form and solve for y:

y - y1 = m(x - x1)

y - 5 = -1/4(x - 4)

y - 5 = -1/4x + 1

y = -1/4x + 1 + 5

y = -1/4x + 6

Proof:

f(x) = -1/4x + 6

f(4) = -1/4(4) + 6

= -1 + 6 = 5, giving (4, 5)

Hope this helps!! Have an Awesome day!!! :-)

4 0

3 years ago

How do you solve this using substitution method 17x+7y=20 Y=20x-42

Nutka1998 [239]

Answer:

17 x + 7 y = 20

y =20x - 42

17 x + 7(20x - 42) = 20

17x + 140x - 294 = 20

157 x = 20 + 294

157x ÷ 157 = 314 ÷ 157

x = 2

y = 20x - 42 

y = 20 (2) - 42

y = 40 - 42

y = -2

there for x = 2 and y = -2 is the solution 

4 0

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

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

X an y are proportional. What is the missing value in the table

Leto [7]

Answer:

X+y

Step-by-step explanation:

6 0

2 years ago