Consider the following sets

Mathematics

2 answers:

Artyom0805 [142]3 years ago

4 0

Answer:

1, 4, 5

Step-by-step explanation:

on edge

choli [55]3 years ago

3 0

Answer:

Paris ∈ F, Abraham Lincoln belongs to none of these sets , The Eiffel Tower is in more than one of these sets.

Step-by-step explanation: :D hope it helps

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What is D to the secind power plus 8D plus 7

Feliz [49]

It matters what D is.
the way to do it is like 8 to the third power th math is:
8 times 8=64
8 times 64=512
8 times 512=4096
3 times

7 0

3 years ago

Solve |p + 2| = 10Solve |p + 2| = 10<br><br> {-12}<br> {-8, 8}<br> {-12, 8}

valentinak56 [21]

By definition, we have

$|p+2| = \begin{cases} p+2 &\text{ if } p+2 \geq 0 \\-p-2 &\text{ if } p+2 < 0 \end{cases}$

So, we have to solve two different equations, depending of the possible range for the variable. We have to remember about these ranges when we decide to accept or discard the solutions:

Suppose that $p+2\geq 0 \iff p \geq -2$

In this case, the absolute value doesn't do anything: the equation is

$p+2 = 10 \iff p = 10-2 = 8$

We are supposing $p \geq -2$ , so we can accept this solution.

Now, suppose that $p+2 < 0 \iff p < -2$ . Now the sign of the expression is flipped by the absolute value, and the equation becomes

$-p-2 = 10 \iff -p = 12 \iff p = -12$

Again, the solution is coherent with the assumption, so we can accept this value as well.

3 0

3 years ago

Which of the following rational functions is graphed below?

vampirchik [111]

Answer:

The correct option is D.

i.e.

$f\left(x\right)=\frac{1}{x\left(x+4\right)}$ is the correct option.

The correct graph is shown in attached figure.

Step-by-step explanation:

Considering the function

$f\left(x\right)=\frac{1}{x\left(x+4\right)}$

$\mathrm{Domain\:of\:}\:\frac{1}{x\left(x+4\right)}\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x$

$\mathrm{Range\:of\:}\frac{1}{x\left(x+4\right)}:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:-\frac{1}{4}\quad \mathrm{or}\quad \:f\left(x\right)>0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-\frac{1}{4}]\cup \left(0,\:\infty \:\right)\end{bmatrix}$