If you pick an integer between 1 and 1,000, what is the probability that it is either divisible by 7 or even (or both?

HELPPPP!!!

Hatshy [7]

Answer:

(x - 1)²/4² - (y - 2)²/2² = 1 ⇒ The bold labels are the choices

Step-by-step explanation:

* Lets explain how to solve this problem

- The equation of the hyperbola is x² - 4y² - 2x + 16y - 31 = 0

- The standard form of the equation of hyperbola is

(x - h)²/a² - (y - k)²/b² = 1 where a > b

- So lets collect x in a bracket and make it a completing square and

also collect y in a bracket and make it a completing square

∵ x² - 4y² - 2x + 16y - 31 = 0

∴ (x² - 2x) + (-4y² + 16y) - 31 = 0

- Take from the second bracket -4 as a common factor

∴ (x² - 2x) + -4(y² - 4y) - 31 = 0

∴ (x² - 2x) - 4(y² - 4y) - 31 = 0

- Lets make (x² - 2x) completing square

∵ √x² = x

∴ The 1st term in the bracket is x

∵ 2x ÷ 2 = x

∴ The product of the 1st term and the 2nd term is x

∵ The 1st term is x

∴ the second term = x ÷ x = 1

∴ The bracket is (x - 1)²

∵ (x - 1)² = (x² - 2x + 1)

∴ To complete the square add 1 to the bracket and subtract 1 out

the bracket to keep the equation as it

∴ (x² - 2x + 1) - 1

- We will do the same withe bracket of y

- Lets make 4(y² - 4y) completing square

∵ √y² = y

∴ The 1st term in the bracket is x

∵ 4y ÷ 2 = 2y

∴ The product of the 1st term and the 2nd term is 2y

∵ The 1st term is y

∴ the second term = 2y ÷ y = 2

∴ The bracket is 4(y - 2)²

∵ 4(y - 2)² = 4(y² - 4y + 4)

∴ To complete the square add 4 to the bracket and subtract 4 out

the bracket to keep the equation as it

∴ 4[y² - 4y + 4) - 4]

- Lets put the equation after making the completing square

∴ (x - 1)² - 1 - 4[(y - 2)² - 4] - 31 = 0 ⇒ simplify

∴ (x - 1)² - 1 - 4(y - 2)² + 16 - 31 = 0 ⇒ add the numerical terms

∴ (x - 1)² - 4(y - 2)² - 16 = 0 ⇒ add 14 to both sides

∴ (x - 1)² - 4(y - 2)² = 16 ⇒ divide both sides by 16

∴ (x - 1)²/16 - (y - 2)²/4 = 1

∵ 16 = (4)² and 4 = (2)²

∴ The standard form of the equation of the hyperbola is

(x - 1)²/4² - (y - 2)²/2² = 1

4 0

3 years ago

Solve by factoring c^2=5c

muminat

Hey there!

c² = 5c

Subtract 5c from both sides:

c² - 5c = 5c - 5c

Simplify :

c² - 5c = 0

c = 5 , c = 0

6 0

2 years ago

Need help on this calculus

N76 [4]

$\displaystyle\int\sin^3t\cos^3t\,\mathrm dt$

One thing you could do is to expand either a factor of $\sin^2t$ or $\cos^2t$ , then expand the integrand. I'll do the first.

You have

$\sin^2t=1-\cos^2t$

which means the integral is equivalent to

$\displaystyle\int\sin t(1-\cos^2t)\cos^3t\,\mathrm dt$

Substitute $u=\cos t$ , so that $\mathrm du=-\sin t\,\mathrm dt$ . This makes it so that the integral above can be rewritten in terms of $u$ as

$\displaystyle-\int(1-u^2)u^3\,\mathrm du=\int(u^5-u^3)\,\mathrm du$

Now just use the power rule:

$\displaystyle\int(u^5-u^3)\,\mathrm du=\dfrac16u^6-\dfrac14u^4+C$

Back-substitute to get the antiderivative back in terms of $t$ :

$\dfrac16\cos^6t-\dfrac14\cos^4t+C$

4 0

2 years ago

Read 2 more answers

(1 point) In this problem we show that the function f(x,y)=7x−yx+y f(x,y)=7x−yx+y does not have a limit as (x,y)→(0,0)(x,y)→(0,0

polet [3.4K]

Answer:

Step-by-step explanation:

Given that,

f(x, y)=7x−yx+y

We want to show that the limit doesn't exist as (x, y)→(0,0).

Limits typically fail to exist for one of four reasons:

1. The one-sided limits are not equal

2. The function doesn't approach a finite value

3. The function doesn't approach a particular value

4. The x - value is approaching the endpoint of a closed interval

a. Considering the case that y=3x

lim(x,y)→(0,0) 7x−yx+y

Since y=3x

lim(x,3x)→(0,0) 7x−3x(x)+3x

lim(x,3x)→(0,0) 10x−3x²

Therefore,

lim(x,3x)→(0,0) 10x−3x² = 0-0=0

b. Let also consider at y=4x

lim(x,y)→(0,0) 7x−yx+y

Since y=4x

lim(x,4x)→(0,0) 7x−4x(x)+4x

lim(x,4x)→(0,0) 11x−4x²

Therefore,

lim(x,4x)→(0,0) 11x−4x² = 0-0=0

c. Let also consider it generally at y=mx

lim(x,y)→(0,0) 7x−yx+y

Since y=mx

lim(x,mx)→(0,0) 7x−mx(x)+mx

lim(x, mx)→(0,0) (7+m)x−mx²

Therefore,

lim(x, mx)→(0,0) (7+m)x−mx² = 0-0=0

But the limit of the given function exist.

So let me assume the function is wrong and the question meant.

f(x, y)= (7x−y) / (x+y)

So, let analyze again

a. Considering the case that y=3x

lim(x,y)→(0,0) (7x−y)/(x+y)

Since y=3x

lim(x,3x)→(0,0) (7x−3x)/(x+3x)

lim(x,3x)→(0,0) 4x/4x

lim(x,3x)→(0,0) 1

Therefore,

lim(x,3x)→(0,0) 1= 1

So the limit is 1

b. Let also consider at y=4x

lim(x,y)→(0,0) (7x−y)/(x+y)

Since y=4x

lim(x,4x)→(0,0) (7x−4x)/(x+4x)

lim(x,4x)→(0,0) 3x/5x

lim(x,4x)→(0,0) 3/5

Therefore,

lim(x,4x)→(0,0) 3/5 = 3/5

So the limit is 3/5

This show that the limit does not exit.

Since one of the condition given above is met, then the limit does not exist. i.e. The function doesn't approach a particular value

c. Let also consider it generally at y=mx

lim(x,y)→(0,0) (7x−y)/(x+y)

Since y=mx

lim(x,mx)→(0,0) (7x−mx)/(x+mx)

lim(x,mx)→(0,0) (7-m)x/(1+m)x

lim(x, mx)→(0,0) (7-m)/(1+m)

Therefore,

lim(x, mx)→(0,0) (7-m)/(1+m) = (7m)/(1+m)

Then, the limit is (7-m)/(1+m)

So the limit doesn't not have a specific value, it depends on the value of m, so the limit doesn't exist.

7 0

3 years ago

Which is the simplified form of 7+5-12?

antiseptic1488 [7]

$\green {\boxed{Answer}}$

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4 0

3 years ago