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2 years ago

Courtney walked 4 miles on Tuesday. How many total feet would that be?

Mathematics

2 answers:

slava [35]2 years ago

6 0

Answer:

21120 ft

Step-by-step explanation:

1 Mile = 5280 Feet

5280*4=21120

Lerok [7]2 years ago

4 0

Answer: 21,120 feet

Step-by-step explanation:

5280(4)=21120

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Tom Johnson worked 32 hours this week for $9.25 per hour. What is Tom’s gross pay for the week?

Dominik [7]

Answer:

296 dollars

Step-by-step explanation:

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2 years ago

I'm so stuck plz help me!!!

elena-14-01-66 [18.8K]

Cd is equal to this: first calculate AB by sin30 which is radical 3 then by sin45 you know bc is radical 3 too then by fisaghures ( idk what you call it)
calculate bd at the end you just gotta do this: bd-bc

$3 - \sqrt{3}$

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2 years ago

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Lim x-> vô cùng ((căn bậc ba 3 (3x^3+3x^2+x-1)) -(căn bậc 3 (3x^3-x^2+1)))

NNADVOKAT [17]

I believe the given limit is

$\displaystyle \lim_{x\to\infty} \bigg(\sqrt[3]{3x^3+3x^2+x-1} - \sqrt[3]{3x^3-x^2+1}\bigg)$

Let

$a = 3x^3+3x^2+x-1 \text{ and }b = 3x^3-x^2+1$

Now rewrite the expression as a difference of cubes:

$a^{1/3}-b^{1/3} = \dfrac{\left(a^{1/3}-b^{1/3}\right)\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)}{\left(a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}\right)} \\\\ = \dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}$

Then

$a-b = (3x^3+3x^2+x-1) - (3x^3-x^2+1) \\\\ = 4x^2+x-2$

The limit is then equivalent to

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}}$

From each remaining cube root expression, remove the cubic terms:

$a^{2/3} = \left(3x^3+3x^2+x-1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3}$

$(ab)^{1/3} = \left((3x^3+3x^2+x-1)(3x^3-x^2+1)\right)^{1/3} \\\\ = \left(\left(x^3\right)^{1/3}\right)^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1x\right)\left(3-\dfrac1x+\dfrac1{x^3}\right)\right)^{1/3} \\\\ = x^2 \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3}$

$b^{2/3} = \left(3x^3-x^2+1\right)^{2/3} \\\\ = \left(x^3\right)^{2/3} \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3} \\\\ = x^2 \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}$

Now that we see each term in the denominator has a factor of x ², we can eliminate it :

$\displaystyle \lim_{x\to\infty} \frac{4x^2+x-2}{a^{2/3}+(ab)^{1/3}+b^{2/3}} \\\\ = \lim_{x\to\infty} \frac{4x^2+x-2}{x^2 \left(\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}\right)}$

$=\displaystyle \lim_{x\to\infty} \frac{4+\dfrac1x-\dfrac2{x^2}}{\left(3+\dfrac3x+\dfrac1{x^2}-\dfrac1{x^3}\right)^{2/3} + \left(9+\dfrac6x-\dfrac1{x^3}+\dfrac4{x^4}+\dfrac1{x^5}-\dfrac1{x^6}\right)^{1/3} + \left(3-\dfrac1x+\dfrac1{x^3}\right)^{2/3}}$

As x goes to infinity, each of the 1/x ⁿ terms converge to 0, leaving us with the overall limit,

$\displaystyle \frac{4+0-0}{(3+0+0-0)^{2/3} + (9+0-0+0+0-0)^{1/3} + (3-0+0)^{2/3}} \\\\ = \frac{4}{3^{2/3}+(3^2)^{1/3}+3^{2/3}} \\\\ = \frac{4}{3\cdot 3^{2/3}} = \boxed{\frac{4}{3^{5/3}}}$

8 0

2 years ago

Dill and Bromberg, Problem 1.23. The San Francisco football team plays better in fair weather. They have a 70% chance of winning

shtirl [24]

Answer:

a) 40% probability that San Francisco will win.

b) 80% probability that the weather was bad

Step-by-step explanation:

We have these following probabilities:

60% chance of snow(bad weather).

If there is bad weather, 20% probability of SF winning.

100-60 = 40% chance of good weather.

If there is good weather, 70% probability of SF winning.

(a) If they play in the Super Bowl in Wisconsin and the weatherman predicts a 60% chance of snow that day, what is the probability that San Francisco will win.

20% of 60%, plus 70% of 40%. So

0.2*0.6 + 0.7*0.4 = 0.4

40% probability that San Francisco will win.

(b) Given that San Francisco lost, what is the probability that the weather was bad?

We use the conditional probability formula to solve this question.

Conditional probability:

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: SF losing

Event B: Bad weather

P(A)

From a), 40% probability that SF wins. So 100-40 = 60% = 0.6 probability that SF losses.

So $P(A) = 0.6$

P(A and B)

60% chance of snow. If it snows, 80% probability of SF losing. So

$P(A \cap B) = 0.6*0.8 = 0.48$

$P(B|A) = \frac{0.48}{0.60} = 0.8$

80% probability that the weather was bad

6 0

3 years ago

Factor the expression completely

Kisachek [45]

Answer:

2 ( 2x+3) (2x-3)

Step-by-step explanation:

8x^2-18Factor out the greatest common factor of 2

2( 4x^2 -9)

2 ( (2x)^2 - 3^2)

What is inside the parentheses is the difference of squares

a^2 - b^2 = (a+b)(a-b)

2 ( 2x+3) (2x-3)

6 0

2 years ago

Read 2 more answers