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3 years ago

PLS HELPPP!!!!!! just to be clear c is not correct

Mathematics

1 answer:

mojhsa [17]2 years ago

8 0

Answer: im gonna guess b?

Step-by-step explanation:

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D: (-infinity, 0) U (0, infinity) because you cant have a zero in the bottom of the fraction

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3 years ago

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7<br> Tuesday<br> Find the Greatest Common<br> Factor of 16 and 28:

Anna35 [415]

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The answer is 4.

Step-by-step explanation:

1. find the factors of 16(1,2,4,4,8,16) and of 28(1,2,4,7,14,28)

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2 years ago

In a group of 31 students, 7 study both Art and Biology.

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2 years ago

Sarah was left with 35 pies after her bake sale. Her neighbor tookſ of the remaining pies. Sarah took home the rest. How many pi

aksik [14]

Answer:

Sarah took home 30 pies

Step-by-step explanation:

Given

Total remaining pieces of pies after bake sale left with Sarah = 35

Her neighbor took 1/7 of the remaining pies

Her neighbor took 1/7 * 35 = 5 pies

Hence, pieces of pies left with Sarah = 35 - 5 = 30

8 0

3 years ago

find the coordinates of the point P on the parabola y=1-x^2 with domain 0≤x≤1 that minimize the area of the triangle enclosed by

coldgirl [10]

Let point P be with coordinates $(x_0,y_0).$ Find the equation of the tangent line.

1. If $y=1-x^2,$ then $y'=-2x.$

2. The equation of the tangent line at point P is

$y-y_0=-2x_0(x-x_0).$

Find x-intercept and y-intercept of this line:

when x=0, then $y=y_0+2x_0^2;$
when y=0, then $x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.$

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

$A=\dfrac{1}{2}\cdot (2x_0^2+y_0)\cdot \left(\dfrac{y_0+2x_0^2}{2x_0}\right)=\dfrac{(y_0+2x_0^2)^2}{4x_0}.$

Since point P is on the parabola, then $y_0=1-x_0^2$ and

$A=\dfrac{(1-x_0^2+2x_0^2)^2}{4x_0}=\dfrac{(1+x_0^2)^2}{4x_0}.$

Find the derivative A':

$A'=\dfrac{2(1+x_0^2)\cdot 2x_0\cdot 4x_0-4(1+x_0^2)^2}{16x_0^2}=\dfrac{12x_0^4+8x_0^2-4}{16x_0^2}.$

Equate this derivative to 0, then

$12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.$

And

$y_0=1-\left(\pm\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{2}{3}.$

Answer: two points: $P_1\left(-\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right), P_2\left(\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right).$

6 0

2 years ago

PLS HELPPP!!!!!! just to be clear c is not correct ​

PLS HELPPP!!!!!! just to be clear c is not correct