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natali 33 [55]
3 years ago
12

Which of the following relations describes a function?

Mathematics
1 answer:
dem82 [27]3 years ago
4 0

Answer:

D. { (-2, -3), (-3, -2), (2, 3), (3, 2) }

Step-by-step explanation:

the x values don't repeat

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Answer this question ASAP
Olin [163]

Answer: start at -3 on the graph then go up one to the right 4

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
. Using the Binomial Theorem explicitly, give the 15th term in the expansion of (-2x + 1)^19
timofeeve [1]
Let's rewrite the binomial as:
(1 - 2x)^{19}

\text{Binomial expansion:} (1 + x)^{n} = \sum_{r = 0}^n\left(\begin{array}{ccc}n\\r\end{array}\right) (x)^{r}

Using the binomial expansion, we get:
\text{Binomial expansion: } (1 - 2x)^{19} = \sum_{r = 0}^{19}\left(\begin{array}{ccc}19\\r\end{array}\right) (-2x)^{r}

For the 15th term, we want the term where r is equal to 14, because of the fact that the first term starts when r = 0. Thus, for the 15th term, we need to include the 0th or the first term of the binomial expansion.

Thus, the fifteenth term is:
\text{Binomial expansion (15th term):} \left(\begin{array}{ccc}19\\14\end{array}\right) (-2x)^{14}
3 0
4 years ago
Use the general solution to solve 5 – 6x = 8x + 17.
Jlenok [28]
<span> 5 – 6x = 8x + 17 
</span>⇔ 5 - 17 = 8x + 6x
⇔ -12 = 14x
⇔ x = -6/7 
7 0
3 years ago
Read 2 more answers
Suppose that y varies inversely with x. Write an equation for the inverse variation. y = 2 when x = 3
sattari [20]
C would be your answer because inverse variation is y=k/x
8 0
3 years ago
Help thankss <br>answer is <br><img src="https://tex.z-dn.net/?f=%20%28%20-%2014.0%29" id="TexFormula1" title=" ( - 14.0)" alt="
lyudmila [28]
We have point P=(x_0,y_0)=(2,-4), so first calculate f'(x_0). There will be:

y=f(x)=2x^3-5x^2\\\\\\\\f'(x)=\big(2x^3-5x^2\big)'=\big(2x^3\big)'-\big(5x^2\big)'=2\big(x^3\big)'-5\big(x^2\big)'=\\\\=2\cdot3x^2-5\cdot2x=\boxed{6x^2-10x}\\\\\\\\f'(x_0)=f'(2)=6\cdot2^2-10\cdot2=6\cdot4-20=24-20=\boxed{4}

Now, we can write the equation of the normal line as:

y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\ y-(-4)=-\dfrac{1}{4}(x-2)\\\\\\y+4=-\dfrac{1}{4}x+\dfrac{1}{2}\\\\\\y=-\dfrac{1}{4}x+\dfrac{1}{2}-4\\\\\\\boxed{y=-\dfrac{1}{4}x-\dfrac{7}{2}}

Normal line (and every line) crosses x-axis when y = 0, so coordinates of A:

\boxed{y=0}\qquad\qquad\text{and}\\\\\\y=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\&#10;0=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\\dfrac{1}{4}x=-\dfrac{7}{2}\quad|\cdot4\\\\\\x=-\dfrac{7\cdot4}{2}\\\\\\x=-7\cdot2\\\\\\\boxed{x=-14}\\\\\\\boxed{A=(-14,0)}
8 0
3 years ago
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