All categories

3 years ago

If a student does not like mounds what is the probability that the student also does not like beach?

Mathematics

1 answer:

Gre4nikov [31]3 years ago

8 0

Try to call an expert

You might be interested in

K-8=-17 that is the answer to this problem

Margarita [4]

Answer:

k=-9

Step-by-step explanation:

k-8+8=-17+8

k=-9

6 0

2 years ago

Read 2 more answers

What is the solution to the equation below? <br> Vx + 3 = x - 3 <br> A. 3<br> B. 6<br> C. 5<br> D. 4

kherson [118]

Answer:

Step 1: Solve Square Root

Vx+3=x-3

x+3=(x-3)^2 (squared both sides)

x+3=x^2-6x+9

x+3-(x^2-6x+9)=0

(-x+1)(x-6)=0 (factor left side of equation)

-x+1=0 or x-6=0

x=1 or x=6

When you plug it in to check

1 (Doesn't Work)

6 (Work)

Therefore, 6 is your solution.

6 0

3 years ago

A tuition bill for school states that Brenda owes $438. She wants to pay off her bill in 6 months. Write and solve an inequality

Marta_Voda [28]

Answer:

Step-by-step explanation:

3 0

3 years ago

PLEASE HELP ASAP!! WILL GIVE BRAINIEST + 30 POINTS!!

Goryan [66]

Given rectangle RUTS, the missing reasons that justifies the five statements in the two-column proof are:

Given
Definition of rectangle.
Definition of rectangle.
By SAS Congruence Theorem.
By CPCTC.

<h3>What is a Rectangle?</h3>

A rectangle is a quadrilateral.
All four angles in a rectangle are right angles.
The opposite sides of a rectangle are parallel and congruent to each other.

Therefore, based on what we are given and the definition of a rectangle, we can establish that △URS ≅ △STU by SAS.

Since △URS ≅ △STU, therefore ∠USR = ∠SUT by CPCTC.

In conclusion, given rectangle RUTS, the missing reasons that justifies the five statements in the two-column proof are:

Given
Definition of rectangle.
Definition of rectangle.
By SAS Congruence Theorem.
By CPCTC.

Learn more about properties of rectangle on:

brainly.com/question/2835318

6 0

2 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

3 years ago