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castortr0y [4]
3 years ago
9

What values do x and y have if /x/ =16 /y/=16 and when x and y are combined they eqaul zero? Explain your reasoning

Mathematics
2 answers:
zaharov [31]3 years ago
6 0

Answer:

When x is 16, then y must be -16.

When x is -16 then y must be 16.

Step-by-step explanation:

We are asked to find the value of x and y for the given condition.

|x|=16

|y|=16

By the definition of absolute value |u|=a, then u=-a\text{ or }u=a, we will get:

x=-16\text{ or }x=16

y=-16\text{ or }y=16

We are also told that when x and y are combined they equal zero.

This means when x is 16, then y must be -16 and when x is -16 then y must be 16.

Ivanshal [37]3 years ago
3 0

Answer:

They have opposite signs. if x=16 then y should be -16 for their sum to be zero.

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V = 1/3 bh for b the base of the cone
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\huge\boxed{b=\dfrac{3V}{h}}

Step-by-step explanation:

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3 years ago
1. Is g = 2 a solution (the correct<br> answer) to 3g = 6? PROVE why or why<br> not.
kherson [118]

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yes this is true

Step-by-step explanation:

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Write -4j - 1 - 4j + 6 in simplest form
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<span><span>
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7 0
2 years ago
Five thousand tickets are sold at​ $1 each for a charity raffle. Tickets are to be drawn at random and monetary prizes awarded a
aksik [14]

Answer:

the expected value of this raffle if you buy 1​ ticket = -0.65

Step-by-step explanation:

Given that :

Five thousand tickets are sold at​ $1 each for a charity raffle

Tickets are to be drawn at random and monetary prizes awarded as​ follows: 1 prize of ​$500​, 3 prizes of ​$300​, 5 prizes of ​$50​, and 20 prizes of​ $5.

Thus; the amount and the corresponding probability can be computed as:

Amount                            Probability

$500 -$1 = $499                1/5000

$300 -$1 = $299                3/5000

$50 - $1 = $49                     5/5000

$5 - $1 = $4                      20/5000

-$1                                   1- 29/5000 = 4971/5000

The expected value of the raffle if 1 ticket is being bought is  as follows:

E(x) = \sum x  * P(x)

E(x) = (499 * \dfrac{1}{5000} + 299 *\dfrac{3}{5000} + 49 *\dfrac{5}{5000} + 4 * \dfrac{20}{5000}  + (-1 * \dfrac{4971}{5000} ))

E(x) = (0.0998 + 0.1794+0.049 + 0.016  + (-0.9942 ))

E(x) = (0.3442 -0.9942 )

\mathbf{E(x) = -0.65}

Thus; the expected value of this raffle if you buy 1​ ticket = -0.65

7 0
3 years ago
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