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Nostrana [21]
3 years ago
5

What is the end behavior of the following Polynomial ? P(x)=-3x^4+3x-2 *

Mathematics
1 answer:
dangina [55]3 years ago
7 0

Answer:

up, down

Step-by-step explanation:

p(x) =  - 3 {x}^{4}  + 3x - 2

It is an even-degree polynomial. Therefore, the start of domain is the opposite the end of domain.

Meaning if the graph starts by decreasing, the graph will end by increasing.

Because the coefficient of highest degree is in negative. Therefore, the graph starts from negative infinity, increasing. The graph will end in positive infinity but decreasing.

Therefore, the answer is first choice.

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I am ab odd number i am less than 100 the sum of my digits is 12 i am a multible of 15 what number amm i
Ivan

Answer:

75

Step-by-step explanation:

7 + 5 = 12

15 x 5 = 75

3 0
3 years ago
PLZ HELP DUE TODAY IN 10 MIN
ICE Princess25 [194]

Answer:

y= 5x + 4

Step-by-step explanation:

slope ontercept

6 0
3 years ago
Read 2 more answers
I need help answering the question
Troyanec [42]

x = 15 + 4y that is what I got out of it

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3 years ago
can someone please help me with this question!!
valkas [14]

Answer:

B' (-4, -4)

Step-by-step explanation:

When you dilate from the origin, you multiply the coordinate points by the scale factor.

B (-1, -1) × 4 = B' (-4, -4)

I hope this helps :))

7 0
3 years ago
21. In 4 + In(4x - 15) = In(5x + 19)
Alexxx [7]

Answer:

x=-30;\quad \:I\ne \:0

Step-by-step explanation:

In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)

\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\

\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\

\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:

\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0

3 0
3 years ago
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