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MaRussiya [10]
3 years ago
5

Which equation describes Which equation describes a line perpendicular to line R that passes through the point (3, -6)?

Mathematics
1 answer:
kompoz [17]3 years ago
4 0

Answer:

B

Step-by-step explanation:

Big

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What is the maximum number of relative extrema contained in the graph of this function f(x)=3x^3-x^2+4x-2
Sergeeva-Olga [200]
The maximum number of turning points in a cubic function is 2.

In this case,

f(x)=3x^3-x^2+4x-2\implies f'(x)=9x^2-2x+4

The discriminant is (-2)^2-4(9)(4)=-140, which means the derivative has no real roots. This means there are no critical points and thus no turning points/relative extrema.
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3 years ago
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(18x3-3x2++x-1)/(x2-4)
Gwar [14]
18x^5-3x^4+x^3-x^2-72x^3+12x^2-4x+4, and then you will combine like terms to be :

18x^5-3x^4-71x^3+11x^2-4x+4
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3 years ago
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11 7/8 + 3 1/5 - ___ = 15
damaskus [11]

Answer: -4.775

Step-by-step explanation: 11(7/8) + 3(1/5) - (-4.775) = 15

5 0
3 years ago
WRITE AN DIVISION PROBLEM THAT HAS A 3 DIGIT DIVIDEND AND DIVISOR BETWEEN 10 AND 20 . SHOW HOW TO SOLVE IT.
Tema [17]
150/15=10
because 10 times 15 equals 150
3 digit dividend = 150
divisor between 10 and 20 is 15
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3 years ago
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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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