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4 years ago

6

Timed! help please! find the value of x.

1 answer:

just olya [345]4 years ago

7 0

Since the square is 90 degress, its equal for each side. X would be parallel to 3 so the answer is 3. :)

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Can someone help with this?

katovenus [111]

Answer:

I havent done this in a long time but I think it is 0x+4=y

Step-by-step explanation:

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3 years ago

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Write twenty-three hundreds as a decimal

mina [271]

Answer:

0.023

Step-by-step explanation:

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4 years ago

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Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt.

qwelly [4]

Answer:

A)

$\displaystyle \frac{dy}{dt}=-\frac{33}{8}$

B)

$\displaystyle \frac{dx}{dt}=\frac{3}{2}$

Step-by-step explanation:

<em>x</em> and <em>y</em> are differentiable functions of <em>t, </em>and we are given the equation:

$xy=6$

First, let's differentiate both sides of the equation with respect to <em>t</em>. So:

$\displaystyle \frac{d}{dt}\left[xy\right]=\frac{d}{dt}[6]$

By the Product Rule and rewriting:

$\displaystyle \frac{d}{dt}[x(t)]y+x\frac{d}{dt}[y(t)]=0$

Therefore:

$\displaystyle y\frac{dx}{dt}+x\frac{dy}{dt}=0$

A)

We want to find dy/dt when <em>x</em> = 4 and dx/dt = 11.

Using our original equation, find <em>y</em> when <em>x</em> = 4:

$\displaystyle (4)y=6\Rightarrow y=\frac{3}{2}$

Therefore:

$\displaystyle \frac{3}{2}\left(11\right)+(4)\frac{dy}{dt}=0$

Solve for dy/dt:

$\displaystyle \frac{dy}{dt}=-\frac{33}{8}$

B)

We want to find dx/dt when <em>x</em> = 1 and dy/dt = -9.

Again, using our original equation, find <em>y</em> when <em>x</em> = 1:

$(1)y=6\Rightarrow y=6$

Therefore:

$\displaystyle (6)\frac{dx}{dt}+(1)\left(-9)=0$

Solve for dx/dt:

$\displaystyle \frac{dx}{dt}=\frac{3}{2}$

5 0

3 years ago

HELP PLEASE!!!!!

Varvara68 [4.7K]

Maybe you should ave a tutor and not ask for answers???

8 0

4 years ago

Explain why P(A|D) and P(D|A) from the table below are not equal.

trasher [3.6K]

Answer:

The Probability that Event A and Event D occur is equal to the probability Event A occurs times the probability that Event D occurs, given that A has occurred.

$P(A\cap D)=P(A)\cdot P(D/A)$

We can find the values of $P(A/D)$ and $P(D/A)$ using the above form formula.

$P(D/A)=\frac{P(A\cap D)}{P(A)}$ ;

$P(A/D)=\frac{P(D\cap A)}{P(D)}$

From the given table, we have the values of P(A), P(D), $P(D\cap A)$ and $P(A\cap D)$ .

Since, Probability= $\frac{The number of wanted outcomes }{the number of possible outcomes}$

∴ $P(A)=\frac{8}{17}$ , $P(D)=\frac{10}{17}$ , $P(A\cap D)=\frac{2}{17}$ and $P(D\cap A)=\frac{2}{17}$

Now, putting these values in above formula we get,

$P(D/A)=\frac{\frac{2}{17}} {\frac{8}{17}}$

$P(D/A)=\frac{2} {8}=\frac{1}{4}$

$P(D/A)= \frac{1}{4}$ .

$P(A/D)=\frac{\frac{2}{17}} {\frac{10}{17}}=\frac{2}{10}$

$P(A/D)=\frac{1}{5}$

As, you can see above that the values of P(A|D) and P(D|A) are not equal.

5 0

3 years ago

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