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3 years ago

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2. From the adjoining Venn-diagram, find the cardinal numbers of the following sets. a) n (U) b) n (A) c) n (B) d) n (AUB) e) n

( AB) f) n (AUB) g) n (A) h) n (B) i) n. (A)

1 answer:

Elena L [17]3 years ago

6 0

as the world caves in :)

step by step explanation:

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Plz help me plzzzzz asap

Varvara68 [4.7K]

Answer:

2 $\frac{2}{9}$

Step-by-step explanation:

First, we will turn 5 2/5 into an improper fraction:

5 * 5 = 25 + 2 = 27/5

So they worked for 27/5 days.

We want to divide the days worked by the distance to get the unit rate of distance per day:

$\frac{12}{\frac{27}{5} } = \frac{12}{1} * \frac{5}{27}=\frac{60}{27}=\frac{20}{9}$

So our answer is 2 $\frac{2}{9}$

Hope this helps, have a great day! :D

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2 years ago

HEEELLLLPPP!!!! PLZZZZ!!!! THANK U!!!<br> (5 screenshots included)

IRISSAK [1]

1.=(2/3,0) 2.=(-3,0) 3.=-10 4.=need a graphing calc 5.=need a graphing calc

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2 years ago

Find the height of a parallelogram with an area of 56 and a base of 20.

qaws [65]

The correct answer is 2.8

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3 years ago

Lily paid her parents $250 for rent and divided the rest of her paycheck evenly into a checking account, a savings account, and

Assoli18 [71]

What kind of equation

6 0

3 years ago

EXPLAIN why we placed the value of x= 4/3( the minimum value) into the equ of gradient(dy/dx) [in the answer, marking scheme att

aliina [53]

$y=x(x-2)^2$
$\implies y'=(x-2)^2+2x(x-2)=3x^2-8x+4=(3x-2)(x-2)=0$
$\implies x=\dfrac23,x=2$

are the critical points, and judging by the picture alone, you must have $b=\dfrac23$ and $a=2$ . (You might want to verify with the derivative test in case that's expected.)

Then the shaded region has area

$\displaystyle\int_0^2x(x-2)^2\,\mathrm dx=\dfrac43$

I'll leave the details to you.

Now, for part (iv), you're asked to find the minimum of $\dfrac{\mathrm dy}{\mathrm dx}=y'$ , which entails first finding the second derivative:

$y'=3x^2-8x+4$
$\implies y''=6x-8$

setting equal to 0 and finding the critical point:

$6x-8=0\implies x=\dfrac86=\dfrac43$

This is to say the minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ *occurs when $x=\dfrac43$ *, but this is not necessarily the same as saying that $\dfrac43$ is the actual minimum value.

The minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ is obtained by evaluating the derivative at this critical point:

$m=\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=4/3}=3\left(\dfrac43\right)^2-8\left(\dfrac43\right)+4=-\dfrac43$

4 0

3 years ago