All categories

3 years ago

can two different numbers have the same absolute value? if yes, give an example. if no, explain why not.

1 answer:

4 0

Yes. Absolute values are always positive, so for example:

the absolute value of 1 is 1, and the absolute value of -1 is also 1

or the absolute value of 7 is 7, and the absolute value of -7 is also 7

and so on!

You might be interested in

<img src="https://tex.z-dn.net/?f=%5Cunderline%7B%5Cboxed%7B%5Cblue%7B%5CLarge%7B%5Cbf%7BChallenge%7D%7D%7D%7D%7D" id="TexFormul

Agata [3.3K]

Step-by-step explanation:

Given point are: P(at², 2at)

Q(a/t², 2a/t)

S(0, 0)

Now, the distance between A(x₁y₁) and B(X₂, y₂)

then AB = √{(x₂ - x₁)² + (y₂ - y₁)²} units

(i) The distance between S and P:

(x₁, y₁) = (0, 0) ⇛x₁= 0, y₁ = 0

(x₂, y₂ ) = (at², 2at) ⇛x₂ = at², y₂ = 2at

SP = √{at² - 0)² + (2at - 0)²}

= √{(at²)² + (2at)²}

= √{a²t²*² + 4a²t²}

= √{a²t⁴ + 4a²t²}

= √{a²t²(t²+4)}

SP = at√(t² + 4)→→→Eqn(1)

(ii) The distance between S and Q :

(x₁, y₁) = (0, 0) ⇛x₁= 0, y₁ = 0

(x₂, y₂ ) = (a/t², 2a/t) ⇛x₂ = a/t², y₂ = 2a/t

SQ = √[{(a/t²) - 0} + {(2a/t) - 0}

= √{(a/t²)² + (2a/t)²}

= √{(a²/t²*²) + (4a²/t²)}

= √{(a²/t⁴) + (4a²/t²)}

= √{(a² + 4a²t²)/t⁴}

= √[{a²(1 + 4t²)}/t⁴]

SQ = (a/t²)√(1 + 4t²) →→→ Eqn(2)

Now,

(1/SP) + (1/SQ) = [1/{at√(t² + 4)}] + [1/{(a/t²)√(1 + 4t²)}]

= (1/at)[1/{√(t² + 4)}] + (t²/a)[1/{(√1 + 4t²)}]

= (1/a)[[1/{t√(t² + 4)}] + [t²/{√(1 + 4t²)}]]

(1/SP) + (1/SQ) = 1/a is not independent of 't'

If suppose S = (a, 0) then

SP = √{(at² - a)² + (2at - 0)²}

= √{a²(t² - 1)² + (2at)²}

= a√{(t² - 1)² + 4t²}

= a√{(t² + 1)²}

SP = a(t² + 1)

1/SP = 1/{a(t² + 1)} →→→Eqn(1)

And

SQ = √[{(a/t²) - a}² + {(2a/t) - 0}²]

= √[a²{(1/t²) - 1}² + a²(2/t)²]

= a√[{(1 - t²)²/t⁴} + (4/t²)]

= a√[{(1 - t²)² + 4t²}/t⁴]

( a/t²)√(1 + t²)²

SQ = (a/t²)(1 + t²)

1/SQ = 1/{(a/t²)(1 + t²)} = t²/{a(1 + t²)} →→→Eqn(2)

Therefore, (1/SP) + (1/SQ)

= 1/{a(t² + 1)} + t²/{a(1 + t²)}

= (1 + t²)/a(1 + t²)

= 1/a

1/a is independent of 't'.

6 0

2 years ago

Which ratios are equivalent to the given ratio?

Alex73 [517]

I wanna say 1:4 hope this helped

5 0

3 years ago

Read 2 more answers

A bowl contains 6 green grapes, 10 red grapes, and 8 black grapes. Which of the following is the correct

irina1246 [14]

Using the it's concept, the correct calculation for the probability of choosing a red grape and then without putting the red grape back into the bowl, choosing a green grape is:

$p = \frac{5}{2} \times \frac{1}{23} = \frac{5}{46}$

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In this problem, we have that there is a total of 6 + 10 + 8 = 24 grapes. For the first grape, 10 are red, hence the probability that the first grape is red is:

$pR = \frac{10}{24} = \frac{5}[12}$

For the second grape, considering that the first was red and was removed, there will be 23 grapes, out of which 6 will be green, hence the probability that the second grape is green is:

$pG = \frac{6}{23}$