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GuDViN [60]
3 years ago
6

NEED HELP ASAP PLEASE

Mathematics
2 answers:
ki77a [65]3 years ago
5 0
D.) is the right answer
OverLord2011 [107]3 years ago
3 0
D is right as the conditions suit to the given situation
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What is the probability of getting either a sum of 66 or at least one 66 in the roll of a pair of​ dice?
Sindrei [870]
So I'm hoping you mean 6, or else the probability is 0. So the possible ways of getting a sum of 6 is 1&5, 2&4, 3&3, 4&2, 5&1. There are 36 different ways of rolling the dice, so divide the 5 possible ways by 36. Then, we need to add the possibility of getting at least one 6 in the roll. It doesn't matter what the other die is, so for each die, the probability is 1/6, and for both, the probability is 2/6. Add this to the probability of getting a sum of 6, and you get your answer.
7 0
3 years ago
Which is closer to 7?
Anna [14]

Answer:

A. 5.3

Step-by-step explanation:

While both of these numbers are close to 7, 5.3 is the closest and we can find this out with subtraction. So, 9.9 - 7 = 2.9, but 7 - 5.3 = 1.7. Which one can we determine is closer? Well, 2.9 > 1.7, so our answer must be A. 5.3!

Have a nice day and I hope this helped!

8 0
3 years ago
Read 2 more answers
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
What is the prime factorization of 28? Please Help! This is due tomorrow! I really want to get an A! Help, please!!!!!!!!!!!!
vovangra [49]
The answer is 2•2•7
Hope this helps
4 0
3 years ago
What is the sum of 8 1/2 + 6 3/8
lyudmila [28]
<span>8 1/2 + 6 3/8
=</span><span>8 4/8 + 6 3/8
= 14 7/8</span>
8 0
3 years ago
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