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2 years ago

7

Consider the function y=f(x)=3x. The values of f(1/2) and f(1/4), rounded to the nearest hundredth, are_ and____ , r

espectively.

1 answer:

puteri [66]2 years ago

3 0

Answer:

f(1/2)=1.5

f(1/4)=0.75

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Show that in any group of 101 people there are at least two persons having the same number friends (It is assumed that if a pers

vladimir1956 [14]

Answer:

Step-by-step explanation:

Let us assume that if possible in the group of 101, each had a different number of friends. Then the no of friends 101 persons have 0,1,2,....100 only since number of friends are integers and non negative.

Say P has 100 friends then P has all other persons as friends. In this case, there cannot be any one who has 0 friend. So a contradiction. Hence proved

Part ii: EVen if instead of 101, say n people are there same proof follows as

if different number of friends then they would be 0,1,2...n-1

If one person has n-1 friends then there cannot be any one who does not have any friend.

Thus same proof follows.

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

Deshaun and his wife are each starting a saving plan. Deshaun will initially set aside $750 and then add $135 every month to the

Bumek [7]

If A=750+135N; and B=485N, then A+B=
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2 years ago

Read 2 more answers

area of rectangular field is 726 sq m if the ratio between its length and breadth is 3:2' find the length and breadth.

ale4655 [162]

Answer:

length = 3x = 3 × 11 = 33 m

breadth = 2x = 2 × 11 = 22 m

Step-by-step explanation:

here's the solution :-

let's take the ratio constants be x

so, length = 3x

breadth = 2x

And, we know that area of rectangle =

length × breadth

So,

=》

$3x \times 2x = 726$

=》

$6 {x}^{2} = 726$

=》

${x}^{2} = \frac{726}{6}$

=》

${x}^{2} = 121$

=》

$x = \sqrt{121}$

=》

$x = 11$

so, length = 3x = 3 × 11 = 33 m

breadth = 2x = 2 × 11 = 22 m

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2 years ago

I need help with geometry

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9514 1404 393

Answer:

A

Step-by-step explanation:

Attached is a copy of the figure with some of the clutter removed. The congruent angles marked with a single arc, and the parallel lines are given.

The angles marked with a double arc are congruent because they are alternate interior angles where transversal XZ crosses parallel lines XW and YZ. Of course, side XZ is congruent to itself.

Hence we have two angles and a side not between them that are congruent. We can declare ∆XWZ ≅ ∆ZYX by the AAS postulate. (choice A)

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2 years ago