Solve: -8 = -7 + x X =

d = {x|x is a whole number} e = {x|x is a perfect square between 1 and 9} f = {x|x is an even number greater than or equal to 2

Alexxandr [17]

$D=\{0;\ 1;\ 2;\ 3;\ 4;...\}\\E=\{4\}\\F=\{2;\ 4;\ 6;\ 8\}\\\\D\ \cap\ F=\{2;\ 4;\ 6;\ 8\}=F\to\boxed{c.}\\conclusion\ F\subset D$

5 0

3 years ago

Read 2 more answers

NEED HELP ASAP PLEASE WITH #9 and #10!!

ICE Princess25 [194]

9) Because the total number of pencils is 180 and you will use them up in 30 days, the equation will have to equal 0 total pencils when 30 is substituted in for the time factor, or x. This already takes our choice 3 because it doesn’t meet this criteria.
The answer to how many pencils in 20 days could be answered by plugging in 20 for x. Choice 4 cannot work because it results in a negative number of pencils. Choices 1 and 2 use the same equation, so by plugging in 20 it is clear choice 2 is the correct answer.

10) A line parallel to the go en equation would have the same slope, -3, which means choices 1 and 4 are out. Plug in (-2,5) into both choices 2 and 3. Plugging -2 into x in choice 2 gives -5, and in choice 3 gives 5 for the y value. Therefore choice 3 is correct.

8 0

3 years ago

1 2/6 as an improper fraction

gladu [14]

Answer:

=4/3

Step-by-step explanation:

Step 1

Multiply the denominator by the whole number

6 × 1 = 6

Step 2

Add the answer from Step 1 to the numerator

6 + 2 = 8

Step 3

Write answer from Step 2 over the denominator

8/6

=86

You can also reduce this fraction.

Find the Greatest Common Factor (GCF) of 8 and 6, if it exists, and reduce our fraction by dividing both numerator and denominator by it,

GCF = 2

=8÷2=4

6/2=3

Simplified Solution

=6×1+2/6 =8/6=4/3

hope this helps

you welcome and thanks

3 0

3 years ago

The rate of transmission in a telegraph cable is observed to be proportional to x2ln(1/x) where x is the ratio of the radius of

sergij07 [2.7K]

Answer:

The value of x that gives the maximum transmission is 1/√e ≅0.607

Step-by-step explanation:

Lets call f the rate function f. Note that f(x) = k * x^2ln(1/x), where k is a positive constant (this is because f is proportional to the other expression). In order to compute the maximum of f in (0,1), we derivate f, using the product rule.

$f'(x) = k*((x^2)'*ln(1/x) + x^2*(ln(1/x)')) = k*(2x\,ln(1/x)+x^2*(\frac{1}{1/x}*(-\frac{1}{x^2})))\\= k * (2x \, ln(1/x)-x)$

We need to equalize f' to 0

k*(2x ln(1/x) - x) = 0 -------- We send k dividing to the other side
2x ln(1/x) - x = 0 -------- Now we take the x and move it to the other side
2x ln(1/x) = x -- Now, we send 2x dividing (note that x>0, so we can divide)
ln(1/x) = x/2x = 1/2 ------- we send the natural logarithm as exp
1/x = e^(1/2)
x = 1/e^(1/2) = 1/√e ≅ 0.607

Thus, the value of x that gives the maximum transmission is 1/√e.

7 0

3 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago

Solve: -8 = -7 + x X =​

Solve: -8 = -7 + x X =