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2 years ago

Pls do some research but can I bring a bearded dragon on a 12 hour car trip without any heath issues happening?

Mathematics

1 answer:

creativ13 [48]2 years ago

5 0

Answer:

after doing a little bit of research. NO your bearded dragon will not be affected as long as you feed it and keep it on a level surface so the enclosure it's in does not tip at all other than that it will be perfectly fine.

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Bodmas 6x (3+8) -8x(6-2)

Leona [35]

If your looking for the answer then it’s 34

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2 years ago

Help!! Not good with Slope! especially with no dots! Brianliest answer offered!!

Georgia [21]

Find two "dots" where the line crosses the X and Y axis at real numbers.

(0,3) and (6,-1)

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3 years ago

I roll a fair die twice and obtain two numbers X1= result of the first roll and X2= result of the second roll. Given that I know

azamat

By definition of conditional probability,

$P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{P((X_1=4\text{ or }X_2=4)\text{ and }X_1+X_2=7)}{P(X_1+X_2=7)}$

$=\dfrac{P((X_1=4\text{ and }X_1+X_2=7)\text{ or }(X_2=4\text{ and }X_1+X_2=7))}{P(X_1+X_2=7)}$

Assuming a standard 6-sided fair die,

if $X_1=4$ , then $X_1+X_2=7$ means $X_2=3$ ; otherwise,
if $X_2=4$ , then $X_1=3$ .

Both outcomes are mutually exclusive with probability $\frac1{36}$ each, hence total probability $\frac2{36}=\frac1{18}$ .

Of the 36 possible outcomes, there are 6 ways to sum the integers 1-6 to get 7:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

and so a sum of 7 occurs $\frac6{36}=\frac16$ of the time.

Then the probability we want is

$P(X_1=4\text{ or }X_2=4\mid X_1+X_2=7)=\dfrac{\frac1{18}}{\frac16}=\frac13$

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2 years ago

How do you solve this I know the answer is x But I want to know how you get to there

Vesnalui [34]

Answer:

Step-by-step explanation:

(–3÷(–3/(x–2)))+2

=((–3×(x–2))÷(–3))+2

=(1×(x–2)÷1)+2

=(x–2)+2

5 0

3 years ago

Let a = x2 + 4. Use a to find the solutions for the following equation:

Zepler [3.9K]

The solutions for x are -2, 0, 2

Solution:

Given that,

$\text { Let } a = x^2 + 4$

Given equation is:

$(x^2 + 4)^2 + 32 = 12x^2 + 48$

$(x^2 + 4)^2 + 32 = 12(x^2 + 4)$

$\text { Subtsitute } a = x^2 + 4 \text{ in above equation }$

$a^2 + 32 = 12a\\\\a^2 -12a + 32 = 0$

$\text{ Let us factorize the above equation }\\\\\text{ Splitting the middle term -12a as -4a - 8a we get, }$

$a^2 -4a - 8a + 32 = 0$

Taking "a" as common term from first two terms and taking "-8" as common from last two terms

$a(a-4)-8(a - 4) = 0$

$\text{Taking (a - 4) as common term, }\\\\(a - 4)(a - 8) = 0$

$\text{Equating to zero we get, }\\\\(a - 4) = 0 \text{ or } (a - 8) = 0\\\\a = 4 \text{ or } a = 8$

$\text{Now substitute the value of a = 8 and a = 4 in: }\\\\a = x^2 + 4$

$\text{ For a = 8: }\\\\a = x^2 + 4\\\\8 = x^2 + 4\\\\x^2 = 4\\\\x = \pm2\\\\x = +2 \text{ or } -2$

$\text{For a = 4: }\\\\a = x^2 + 4\\\\4 = x^2 + 4\\\\x = 0$

$\text{Thus solutions of } x \text{ are: }\\\\x = 0 \text{ or } x = 2 \text{ or } x = -2$

8 0

2 years ago

Pls do some research but can I bring a bearded dragon on a 12 hour car trip without any heath issues happening?​

Pls do some research but can I bring a bearded dragon on a 12 hour car trip without any heath issues happening?