2 years ago

Have you ever seen 2 pretty best friends?

Mathematics

2 answers:

ozzi2 years ago

3 0

Answer:

yea me and myself

Step-by-step explanation:

Korvikt [17]2 years ago

3 0

nah man

my friends and i all ugly

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Relative to the acute angle θ, the three sides of a right triangle are the

Rufina [12.5K]

Opposite/adjacent/hypotenuse

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2 years ago

There are only green yellow blue cubes in a bag

GREYUIT [131]

Answer:

1 ÷ 11

Step-by-step explanation:

The computation is shown below:

let us assume the yellow cubes be x

So, blue cubes be 2x

And, green cubes be 2x (4) 8x

Now the ratio is

Yellow: Blue: green

x:2x:8x

1:2:8

Now the probability of yellow cubes is

= 1 ÷ (1 + 2 + 8)

= 1 ÷ 11

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2 years ago

Tell which of the ordered pairs is a solution to the equation y=1/2x+2 (0,1) (2,2) (4,4) or (-2,1)

Softa [21]

y = 1/2 x+2

y = 0 +2

(0,2)

y = (1/2)* 2 +2

(2,3)

y =1/2*4+2

(4,4) Yes!

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3 years ago

Prove the following statement using a proof by contraposition. Yr EQ,s ER, if s is irrational, then r + 1 is irrational.

sp2606 [1]

Answer:

I think that what you are trying to show is: If $s$ is irrational and $r$ is rational, then $r+s$ is rational. If so, a proof can be as follows:

Step-by-step explanation:

Suppose that $r+s$ is a rational number. Then $r$ and $r+s$ can be written as follows

$r=\frac{p_{1}}{q_{1}}, \,p_{1}\in \mathbb{Z}, q_{1}\in \mathbb{Z}, q_{1}\neq 0$

$r+s=\frac{p_{2}}{q_{2}}, \,p_{2}\in \mathbb{Z}, q_{2}\in \mathbb{Z}, q_{2}\neq 0$

Hence we have that

$r+s=\frac{p_{1}}{q_{1}}+s=\frac{p_{2}}{q_{2}}$

Then

$s=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}\in \mathbb{Q}$

This is a contradiction because we assumed that $s$ is an irrational number.

Then $r+s$ must be an irrational number.

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2 years ago

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Answer: omggg I need that one too irdk I need help

Step-by-step explanation:

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2 years ago

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