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3 years ago

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A universal set contains only sets A and B. If A∩B = Ø and A and B are not the empty set, then all of the following are true exc

ept ______.
A. A and B are disjoint.
B. A is equal to B complement.
C. A∪B={}
D. A is not a subset of B.

2 answers:

DaniilM [7]3 years ago

8 0

B?................................

mafiozo [28]3 years ago

6 0

Answer:The correct answer among the choices provided is the third option. All of the other statements are true except A∪B={}. Both A and B are said to be non-empty. The open and close parentheses is equal to an empty set.

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What is the value of h? g = -7.1 h = 3 - 10g

Step2247 [10]

Answer:

74

Step-by-step explanation:

Please separate the two functions:

g = -7.1

h = 3 - 10g

Substitute -7.1 for g in h = 3 - 10g:

h = 3 - 10(-7.1) = 71 + 3 = 74

4 0

3 years ago

Translate the sentence into an Intequality<br><br><br> Eight subtracted from y is greater than 25

Brums [2.3K]

Answer:

8-y<25 hope this answers it

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3 years ago

A man starts walking north at 2 ft/s from a point P. Five minutes later a woman starts walking south at 3 ft/s from a point 500

Vedmedyk [2.9K]

Answer:

The rate at which both of them are moving apart is 4.9761 ft/sec.

Step-by-step explanation:

Given:

Rate at which the woman is walking, $\frac{d(w)}{dt}$ = 3 ft/sec

Rate at which the man is walking, $\frac{d(m)}{dt}$ = 2 ft/sec

Collective rate of both, $\frac{d(m+w)}{dt}$ = 5 ft/sec

Woman starts walking after 5 mins so we have to consider total time traveled by man as (5+15) min = 20 min

Now,

Distance traveled by man and woman are $m$ and $w$ ft respectively.

⇒ $m=2\ ft/sec=2\times \frac{60}{min} \times 20\ min =2400\ ft$

⇒ $w=3\ ft/sec = 3\times \frac{60}{min} \times 15\ min =2700\ ft$

As we see in the diagram (attachment) that it forms a right angled triangle and we have to calculate $\frac{dh}{dt}$ .

Lets calculate h.

Applying Pythagoras formula.

⇒ $h^2=(m+w)^2+500^2$

⇒ $h=\sqrt{(2400+2700)^2+500^2} = 5124.45$

Now differentiating the Pythagoras formula we can calculate the rate at which both of them are moving apart.

Differentiating with respect to time.

⇒ $h^2=(m+w)^2+500^2$

⇒ $2h\frac{d(h)}{dt}=2(m+w)\frac{d(m+w)}{dt} + \frac{d(500)}{dt}$

⇒ $\frac{d(h)}{dt} =\frac{2(m+w)\frac{d(m+w)}{dt} }{2h}$ ...as $\frac{d(500)}{dt}= 0$

⇒ Plugging the values.

⇒ $\frac{d(h)}{dt} =\frac{2(2400+2700)(5)}{2\times 5124.45}$ ...as $\frac{d(m+w)}{dt} = 5$ ft/sec

⇒ $\frac{d(h)}{dt} =4.9761$ ft/sec

So the rate from which man and woman moving apart is 4.9761 ft/sec.

3 0

3 years ago

A construction crew must build 4 miles of road in one week. On Monday, they build 12 mile of road. On Tuesday, they build 13 mil

Julli [10]

$\frac{19}{6}$ many more miles of road are left to build .

<u>Step-by-step explanation:</u>

Here we have , A construction crew must build 4 miles of road in one week. On Monday, they build 1/2 mile of road. On Tuesday, they build 1/3 mile of road. We need to find that How many more miles of road are left to build . Let's find out:

On Monday, they build 1/2 mile of road.
On Tuesday, they build 1/3 mile of road.

Let Miles of road left to build is x So,

⇒ $\frac{1}{2} +\frac{1}{3} +x = 4$

⇒ $x = 4 - (\frac{1}{2} +\frac{1}{3})$

⇒ $x = 4 - (\frac{1(3)+1(2)}{6} )$

⇒ $x = 4 - (\frac{5}{6} )$

⇒ $x = (\frac{4(6)-5}{6} )$

⇒ $x = \frac{24-5}{6}$

⇒ $x = \frac{19}{6}$

Therefore, $\frac{19}{6}$ many more miles of road are left to build .

5 0

3 years ago

Read 2 more answers

1/2 divided by 3/4 = 1/a times b/c = d/e

Vinvika [58]

Step-by-step explanation:

1/2 ÷ 3/4 = 1/2 × 4/3 = 4/6

a = 2

b = 4

c = 3

d = 4

e = 6

6 0

3 years ago