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3 years ago

Find the GCF of 30x^3 and 12x^4

Mathematics

1 answer:

Allushta [10]3 years ago

7 0

Answer:6x^3

Step-by-step explanation: That’s the biggest factor that can be taken out of both equations

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I REALLY NEED HELP PLEASE

Nezavi [6.7K]

Answer:

Step-by-step explanation:

given

$\frac{s + ( - 7)}{ - 1}$

we first need to simplify the expression before we find it's value when s=-3

positive times negative=negative

so the equation becomes

$\frac{s - 7}{ - 1}$

substituting s=-3 into the equation.

$\frac{ - 3 - 7}{ - 1}$

$\frac{ - 10}{ - 1}$

=the negatives cancel out leaving

$= 10$

7 0

3 years ago

What do all points that lie on the y-axis have in common in terms of their coordinates?

vitfil [10]

For all points that lie on the y-axis, the x-coordinate is zero.

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

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

if alyssa plans to run 7/8 mile on a trail at one point she sees that she is at mile maker 0.3 How much farther does she have to

Sever21 [200]

Rewrite 7/8 as a decimal by dividing 7 by 8:

7/8 = 0.875

Now to find how much further she has to run subtract where is now from the total distance:

0.875 - 0.3 = 0.575

She needs to run 0.575 miles more

6 0

2 years ago

A photographer buys a new camera for $1350. Every year the camera decreases in value by 10%.

bagirrra123 [75]

Answer:

A(t)=1350(0.9)^t

i hope this helps :)

4 0

2 years ago

Find the GCF of 30x^3 and 12x^4​

Find the GCF of 30x^3 and 12x^4