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antiseptic1488 [7]
3 years ago
11

Chelsea mailed 2 boxes and 2 bags to her uncle in army. The two boxes weighed 8 ½ pounds each and the two bags weighed 4 ½ pound

s each. How many pounds do all 2 boxes and 2 bags weigh?
Mathematics
2 answers:
Orlov [11]3 years ago
7 0

Answer:26

Step-by-step explanation:

4.5x2=9+17=26

8.5x2=17

Roman55 [17]3 years ago
5 0

Answer:

26lbs.

Step-by-step explanation:

this problem is quite simple when we break it down

first we multiply each of the amounts of the bags and the boxes by two to represent the two of each items as shown

8 1/2 * 2 = 17

4 1/2 * 2 = 9

then all we have to do is add the two new amounts together and TADA:

17 + 9 = 26

we have our answer

NOTE: when multiplying a fraction, you are multiplying how many parts (the numerator) and NOT the size of the part within the whole number (the denominator) in this case 1/2 * 2 = 2/2 and 2/2 = 1

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What formula would you use to find the area of this figure?
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Step-by-step explanation:

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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
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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).

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