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2 years ago

The absolute value of 18

1 answer:

6 0

The absolute value of 18 is 18. In order to find an absolute value, you take the positive value of whatever the number may be. So the absolute value of 9 and -9 would both still be 9.

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Nine children (Andrew, Ben, Candy, David, Ella, Fiona, Gordon, Henry, Ivan) form a circle in alphabetical order, clockwise. To c

blondinia [14]

We should start counting from fiona

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2 years ago

Consider the function g(x) = -3x - 4 if the domain value is selected from the set {-8, -2, 0, 3}, which of the following is a co

Marina86 [1]

Answer:

C. -13

Step-by-step explanation:

Solve for g(x) by replacing x with each domain value in the set. Solving for g(x) will give you these numbers:

{20, 2, -4, -13}

-13 is the only value that appears in the range set and in the multiple choice solutions. Therefore, the answer is C. -13

7 0

2 years ago

Read 2 more answers

The number 42 is increased to 46. what is the percentage by which the number was increased?

Arte-miy333 [17]

Answer:

9.52% ( to 2 dec. places )

Step-by-step explanation:

Calculate the percentage increase using

percent increase = $\frac{increase}{original}$ × 100%

increase = 46 - 42 = 4, hence

percent increase = $\frac{4}{42}$ × 100% ≠ 9.52%

7 0

2 years ago

Limit question: lim x-->pi ((e^sinx)-1)/(x-pi)

aleksandr82 [10.1K]

$\displaystyle\lim_{x\to\pi}\dfrac{e^{\sin x}-1}{x-\pi}$

Notice that if $f(x)=e^{\sin x}$ , then $f(\pi)=e^{\sin\pi}=e^0=1$ . Recall the definition of the derivative of a function $f(x)$ at a point $x=c$ :

$f'(c):=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

So the value of this limit is exactly the value of the derivative of $f(x)=e^{\sin x}$ at $x=\pi$ .

You have

$f'(x)=\cos x\,e^{\sin x}\implies f'(\pi)=\cos\pi\,e^{\sin\pi}=-1$

4 0

2 years ago

For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?

Bond [772]

Given that $r=\theta$ , then $r'=1$

The slope of a tangent line in the polar coordinate is given by:

$m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$

Thus, we have:

$m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}$

Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

$\sin\theta+\theta\cos\theta=0 \\ \\ \theta\cos\theta=-\sin\theta \\ \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta$

Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

θ = 0

θ = 2.02875783811043

θ = 4.91318043943488

Part B:

For vertical tangent lines, $\frac{1}{m} =0$

Thus, we have:

$\cos\theta-\theta\sin\theta=0 \\ \\ \Rightarrow\theta\sin\theta=\cos\theta \\ \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta$

Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

θ = 4.91718592528713

3 0

2 years ago