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Alenkinab [10]
3 years ago
9

How was y'alls day? :)

Mathematics
2 answers:
VikaD [51]3 years ago
8 0
Not the best, teachers are so annoying lol. Hope you are doing well though, don’t give up❤️
Mrac [35]3 years ago
3 0

answer-good how was yours

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PLEASE ANSWER NUMBER FOUR AND SHOW WORK THANK YOU IN ADVANCE.
Paladinen [302]

Answer:

A,C, AND D

Step-by-step explanation:

A. 3+4>5 therefore a triangle

B. 1+7<10 therefore not a triangle

C. 1.5+1.5>2.5 therefore a triangle

D. 7+7>7 therefore a triangle

E. 6+6=12 therefore not a triangle

8 0
4 years ago
Solve the given initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1
Yuri [45]
Let's check if the ODE is exact. To do that, we want to show that if

\underbrace{(x+y)^2}_M\,\mathrm dx+\underbrace{(2xy+x^2-2)}_N\,\mathrm dy=0

then M_y=N_x. We have

M_y=2(x+y)
N_x=2y+2x=2(x+y)

so the equation is indeed exact. We're looking for a solution of the form \Psi(x,y)=C. Computing the total differential yields the original ODE,

\mathrm d\Psi=\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0
\implies\begin{cases}\Psi_x=(x+y)^2\\\Psi_y=2xy+x^2-2\end{cases}

Integrate both sides of the first PDE with respect to x; then

\displaystyle\int\Psi_x\,\mathrm dx=\int(x+y)^2\,\mathrm dx\implies\Psi(x,y)=\dfrac{(x+y)^3}3+f(y)

where f(y) is a function of y alone. Differentiate this with respect to y so that

\Psi_y=2xy+x^2-2=(x+y)^2+f'(y)
\implies2xy+x^2-2=x^2+2xy+y^2+f'(y)
f'(y)=-2-y^2\implies f(y)=-2y-\dfrac{y^3}3+C

So the solution to this ODE is

\Psi(x,y)=\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3+C=C

i.e.


\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3=C
6 0
3 years ago
Twice a number is at most 36.​
jekas [21]

Answer:

18?

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
The composite figure is made up of a <br> two<br><br> figures<br> on top of a <br><br> .
mars1129 [50]

That’s not a question? I need you to specify on what the question is then I can edit this post and let you know the answer

3 0
3 years ago
4y4 – 5y2 – 9 = 0.<br> how do you solve this question
umka21 [38]

Answer:

-3/2 or 3/2

Step-by-step explanation:

Step 1: Factor left side of equation.

(y2+1)(2y+3)(2y−3)=0

Step 2: Set factors equal to 0.

y2+1=0 or 2y+3=0 or 2y−3=0

6 0
3 years ago
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