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3 years ago

PLEASE HELP ME ASAP! I WILL MARK BRAINLIEST! I NEED HELP!! EXPLAIN!!

Mathematics

1 answer:

notsponge [240]3 years ago

3 0

Following refers to the annotated image:
Imagine yourself looking from the top, then the cube shaded blue will be cube 13. We call this the origin.
After that, follow the 3-D view, move 1 up shades cube 9.
move 1 right gives the cube 10
move another right -> cube 11
move one up -> cube 7
another up -> cube 3
Finally, move one right -> cube 4

Final answer: see image

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A diagram of Tracy’s deck is shown below, shaded blue. He wants to cover the

Lelechka [254]

Hey! We can’t see a picture of the problem.

4 0

3 years ago

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

3 years ago

Please help me with this question

gayaneshka [121]

9514 1404 393

Answer:

Step-by-step explanation:

The leading term is of odd degree and negative coefficient. The odd degree tells you the end behaviors will have opposite signs. The negative coefficient tells you the left end behavior will be positive, and the right end behavior will be negative.

$\textbf{D. }\lim\limits_{x\to -\infty}f(x)=\infty\quad\lim\limits_{x\to \infty}f(x)=-\infty$

5 0

2 years ago

Please help! WILL MARK BRAINLY

kodGreya [7K]

Answer:

D above and to the left

Step-by-step explanation: The origin is in the center of a grid so going left and up will put you in quadrant 2 where x is negative and y is positive

8 0

3 years ago

Annie writes the numbers 1 through 10 on note cards. She flips the cards over so she cannot see the number and selects three car

Kay [80]

<h2>The probability that Annie has selected the cards numbered 1, 2, and 3 is $(\frac{1}{720} )$ </h2>

Step-by-step explanation:

Here, the total number of cards in the set = 10

The cards are numbered as 1,2,3,4,5,...., 10

P(Any event E) = $\frac{\textrm{Total favorable events}}{\textrm{Total number of events}}$

P(picking a card with number 1) = $\frac{\textrm{Total cards with number 1 on it}}{\textrm{Total cards}} = (\frac{1}{10})$

Now when the first card is picked, the number of cards left = 10 - 1 = 9

P(picking a card with number 2) = $\frac{\textrm{Total cards with number 2 on it}}{\textrm{Total cards}} = (\frac{1}{9})$

Similarly, P(picking a card with number 3) = $\frac{\textrm{Total cards with number 3 on it}}{\textrm{Total cards}} = (\frac{1}{8})$

So, the total probability that she has selected card with number 1, 2 and 3

= $(\frac{1}{10} ) \times (\frac{1}{9} ) \times (\frac{1}{8} ) = (\frac{1}{720} )$

Hence, the probability that she has selected the cards numbered 1, 2, and 3 is $(\frac{1}{720} )$

4 0

3 years ago

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