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3 years ago

What is a prime number

Mathematics

2 answers:

creativ13 [48]3 years ago

5 0

A prime number is a number that has only two factors 1 and the number itself

miss Akunina [59]3 years ago

4 0

Answer:

a number that is divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11).

Step-by-step explanation:

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Arrange the numbers in increasing order.

SVEN [57.7K]

Answer:14.200,14.204,14.210,14.240

Step-by-step explanation:

8 0

3 years ago

Multiplying or dividing two related quantites bt the same number is called

stira [4]

it is called dun dun dun this Equivalent Ratio

7 0

3 years ago

Two cards are drawn without replacement from a standard deck of 52 playing cards. What is the probability of choosing a red card

joja [24]

Answer: $\dfrac{3}{51}$

Step-by-step explanation:

Given : The total number of cards in a deck = 52

Number of red cards = 26

There are two types of red cards : diamond and heart.

Number of diamond cards = 13

The probability that the first card is a diamond :-

$\dfrac{13}{52}=\dfrac{1}{4}$

Since diamond is also a red card.

Now, the total cards left = 51

The number of red cards left = 12

The probability that the second card is a red card (without repetition) is given by :-

$\dfrac{12}{51}$

Now, the probability of choosing a red card for the second card drawn, if the first card, drawn without replacement, was a diamond :-

$\dfrac{1}{4}\times\dfrac{12}{51}=\dfrac{3}{51}$

7 0

4 years ago

· The volume of a box in the shape of a cube is 2197 cm3. Find the area of one side of the box.

Yuliya22 [10]

Answer:

Let the side of the cube be x.

Volume = length × width × height

For a cube:

Length = width = height

Now volume = x × x × x = x³

This equals 2197

Therefore :

X³ = 2197

X = ∛2197

= 13

The side of the cube = 13cm

Area = length × width

Length = width = 13cm

Area = 13 × 13 = 169

= 169 cm²

I hope this helps!

4 0

4 years ago

Find K by evaluating Limit

Lubov Fominskaja [6]

Answer:

Step-by-step explanation:

$\lim_{x \to \infty}\frac{1-cos4x}{1-cos2x}= \lim_{x \to 0} \frac{(1-cos4x)'}{(1-cos2x)'}= \lim_{x \to 0}\frac{4sin4x}{2sin2x}= \lim_{x \to 0}\frac{2*2sin2xcos2x}{sin2x}\\ = \lim_{x \to 0}4cos2x\\ =4$

{L'Hospital's rule}

I hope this helps you

:)

6 0

2 years ago