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2 years ago

Can you help me with this?

Mathematics

2 answers:

antiseptic1488 [7]2 years ago

7 0

Answer: C

Step-by-step explanation:

Lady bird [3.3K]2 years ago

5 0

The answer will be B

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What type of association does there appear to be between the variables? explain

Nataliya [291]

Where are the variables for the equation?

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3 years ago

The basketball coach make up a game to play where each game takes. 10 shots at the basket. For every basket made, the player gai

levacccp [35]

Answer:

5s + 5m

5(10) + 5(-15)

50 - 75 = -25 points

Step-by-step explanation:

s = sanked baskets = 10

m = missed baskets = -15

7 0

3 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago

What is the probability of not picking a red face card when you draw a card at random from a pack of 52 cards?

aniked [119]

Answer:

3/13

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

I need help solving a system of linear equation, and it says to solve it by Substitution.

olga_2 [115]

2x - 3y = -24
x+ 6y = 18

you solve the first one by the x:

2x = -24 + 3y

x = (-24+3y)/2

then you substitute what you found in the other equation and you solve by y

(-24+3y)/2 + 6y = 18 multiply everything by 2

-24 + 3y + 12y = 36

3y + 12y = 36 + 24

15y = 60

y = 60/15

y = 4

Good job! You found your y! :D

Now let's substitute the y in the first equation with 4:

2x - 3(4) = -24

2x - 12 = -24

2x = -24 + 12

2x = -12

x = -12/2

x= -6

Done!! ;) <span />

6 0

2 years ago

Read 2 more answers