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Sophie [7]
3 years ago
14

WELPPP !!!!!!!!!!!!!!!!

Mathematics
1 answer:
Vsevolod [243]3 years ago
3 0
The answer is A. To do this you must add 9 to both sides and you get (x+3)^2=27 and you must solve from there
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The quotient of 36 and 3 is j. write expression
Dovator [93]
This might be a mistake but i'll give it a go:

36/3 = j
4 0
3 years ago
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Which shows the prime factorization of 80? Check all that apply. 2 × 4 × 10 2 × 2 × 2 × 2 × 5 24 × 5 2 × 5 × 8
TEA [102]

Answer:

The Prime Factorization of 80 is,  2 × 4 × 10, 2 × 2 × 2 × 2 × 5  and  2 × 4 × 10

Step-by-step explanation: They are correct, because they all equal 80. 2 × 4 × 10=80 and  2 × 4 × 10=80, and 2 × 2 × 2 × 2 × 5=80.

24 × 5=120, Therefore it's the only incorrect question.

4 0
3 years ago
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Describe how you would find the sum of 24,36,and 13.
fenix001 [56]
I would add 24 and 36, to make 60, then I would tack on 13 and get 73.
6 0
3 years ago
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What is the value of x in the equation 2x+3y=36 when y=6
Lady_Fox [76]
2x+3y=36
2x=36-3y
y=6
2x=36-3*6
2x=36-18
2x=18
x=18:2
x=9
7 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
3 years ago
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