What is the range of the graph below?

Mathematics

2 answers:

pickupchik [31]3 years ago

5 0

The answer is C because The x axis always goes first y axis second.

OLga [1]3 years ago

4 0

That’s correct since the lowest y value is -8 and the highest is infinity!

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Evaluate the equation. -4x = 14 A.10 B. 7/2 C. -7/2 D. -10

Alex_Xolod [135]

Answer:

c) -7/2

Step-by-step explanation:

14/-4= -3.5

4 0

3 years ago

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P(x)= 3x^3-5x^2-14x-4

nexus9112 [7]

$\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\
$

$\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)$

$\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$

7 0

3 years ago

Don’t mind the already picked answer..i don’t understand this question though

Tatiana [17]

The answer is = because all of the other dimensions to the triangles are equal.

5 0

3 years ago

A function is a special type of relation where

Citrus2011 [14]

Answer: the answer would be B :)

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3 years ago

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Can someone think of a poem which starts with "Today I feel ...."<br>pls

GuDViN [60]

Yes, I shall give you an idea of how to start it. :)

Today I feel glad...
For all the friends I have had...
They gave me a pleasure...
For all that I treasure...
Special and dear to my heart.

Ta da! :D Hope you like it

7 0

3 years ago

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