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Free_Kalibri [48]

4 years ago

14

(a-a)/(a-a)=2 i cannot solve it. plz heplme

2 answers:

Flura [38]4 years ago

5 0

When a=a
then a-a=0

so
(a-a)/(a-a)=0/0=2?
false
no solution

Katyanochek1 [597]4 years ago

3 0

There is no solution. (a-a) is zero, and division by zero is disallowed.

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Convert the Cartesian equation x^2 + y^2 = 16 to a polar equation.

RideAnS [48]

Answer:

Problem 1: $r=4$

Problem 2: $r=-2\sin(\theta)$

Problem 3: $r\sin(\theta)=3$

Step-by-step explanation:

Problem 1:

So we are going to use the following to help us:

$x=r \cos(\theta)$

$y=r \sin(\theta)$

$\frac{y}{x}=\tan(\theta)$

So if we make those substitution into the first equation we get:

$x^2+y^2=16$

$(r\cos(\theta))^2+r\sin(\theta))^2=16$

$r^2\cos^2(\theta)+r^2\sin^2(\theta)=16$

Factor the $r^2$ out:

$r^2(\cos^2(\theta)+\sin^2(\theta))=16$

The following is a Pythagorean Identity: $\cos^2(\theta)+\sin^2(\theta)=1$ .

We will apply this identity now:

$r^2=16$

This implies:

$r=4 \text{ or } r=-4$

We don't need both because both of include points with radius 4.

Problem 2:

$x^2+y^2+2y=0$

$(r\cos(\theta))^2+(r\sin(\theta))^2+2(r\sin(\theta))=0$

$r^2\cos^2(\theta)+r^2\sin^2(\theta)+2r\sin(theta)=0$

Factoring out $r^2$ from first two terms:

$r^2(\cos^2(\theta)+\sin^2(\theta))+2r\sin(\theta)=0$

Apply the Pythagorean Identity I mentioned above from problem 1:

$r^2(1)+2r\sin(\theta)=0$

$r^2+2r\sin(\theta)=0$

or if we factor out r:

$r(r+2\sin(\theta))=0$

$r=0 \text{ or } r=-2\sin(\theta)$

r=0 is actually included in the other equation since when theta=0, r=0.

Problem 3:

$y=3$

$r\sin(\theta)=3$

8 0

4 years ago

there are 6 times as many ways to arrange the letters in the word "sleeplessness" as there are to arrange the letters in the wor

EastWind [94]

Using arrangements of words, it is found that the statement is true.

------------------

Suppose a word has n letters.
Considering that m of these letters are repeating, each $n_1,n_2,...,n_m$ times.
The number of distinct ways the word can be arranged is given by:

$N = \frac{n!}{n_1\times n_2 \times ... n_m}$

------------------

The word "sleeplessness" has 13 letters.
s repeats 5 times.
e repeats 4 times.
l repeats 2 times.

Thus, the number of ways to arrange the letters is given by:

$N_1 = \frac{13!}{5!4!2!}$

------------------

The word "senselessness" has 13 letters.
s repeats 6 times.
e repeats 4 times.
n repeats 2 times.

Thus, the number of ways to arrange the letters is given by:

$N_2 = \frac{13!}{6!4!2!}$

------------------

Finding the ratio of $N_2$ to $N_1$ :

$\frac{N_2}{N_1} = \frac{\frac{13!}{5!4!2!}}{\frac{13!}{6!4!2!}} = \frac{13!}{5!4!2!} \times \frac{6!4!2!}{13!} = \frac{6!}{5!} = 6$

Thus, since the ratio is 6, the statement is true.

A similar problem is given at brainly.com/question/16790460

8 0

3 years ago

Read 2 more answers

What all are the factors off 453,738

vazorg [7]

Answer:

1, 2, 3, 6, 47, 94, 141, 282, 1609, 3218, 4827, 9654, 75623, 151246, 226869, 453738

Step-by-step explanation:

I looked it up

6 0

3 years ago

Out of the last 50 guests, only 1 of them brought a coupon for free admission to the carnival. If there are 1500 guests expected

Aleonysh [2.5K]

The answer is 30
we expect that in every 50 people, 1 will have free admission.
so:
$1500 \div 50 = 30$

3 0

3 years ago

N 100 mL of a solution there are 36 g. How many mg will 7.5 mL contain?

dexar [7]

Answer:

Answers here just click https://dffrg.com

Step-by-step explanation:

5 0

1 year ago