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3 years ago

How much paper will it take to make each tree including the bottom??

Mathematics

1 answer:

Nataly [62]3 years ago

7 0

The bottom is a square with a side length of 2 ft.

The area of a square is Area = S^2 = 2^2 = 4 square ft. Bottom)

The area of one side ( triangle) = 1/2 x base x height = 1/2 x 2 x 4 = 4 square ft.

There are 4 triangles: 4 x 4 sq. ft. = 16 sq.ft. ( four sides)

Total area = four sides + bottom = 16 + 4 = 20 feet^2

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Given that f(x) = 19x2 + 152, solve the equation f(x) = 0

telo118 [61]

Option A

The solution is:

$x = \pm 2i \sqrt{2}$

Solution:

$f(x) = 19x^2+152$

We have to solve the equation f(x) = 0

Let f(x) = 0

$0=19x^2+152$

Solve the above equation

$19x^2 + 152 = 0$

$\mathrm{Subtract\:}152\mathrm{\:from\:both\:sides}\\\\19x^2+152-152=0-152\\\\Simplify\ the\ above\ equation\\\\19x^2 = -152\\\\\mathrm{Divide\:both\:sides\:by\:}19\\\\\frac{19x^2}{19} = \frac{-152}{19}\\\\x^2 = -8$

Take square root on both sides

$x = \pm \sqrt{-8}\\\\x = \pm \sqrt{-1}\sqrt{8}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\x = \pm i\sqrt{8}\\\\x = \pm i \sqrt{2 \times 2 \times 2}\\\\x = \pm 2i\sqrt{2}$

Thus the solution is found

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3 years ago

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Tcecarenko [31]

Answer:

Step-by-step explanation:

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2 years ago

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Q4.A/Find the linearization L(x, y) of the function f(x, y) = (x + y + 2)² at p. = (1,2)

Dafna11 [192]

Answer:

Find the linearization L(x,y) of the function at each point. f(x,y) = x2 + y2 + 1 a. (4,0) b. (2,0) a. L(x,y) = Find the linearization L(x,y,z) of the function f(x,y,z) = 1x2 + y2 +z2 at the points (7,0,0), (3,4,0), and (4,4,7). The linearization of f(x,y,z) at (7,0,0) is L(x,y,z)= (Type an exact answer, using radicals as needed.)

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3 years ago

Choose three different colors to highlight the Statement in the proof and match it to it's angle or line segment in the diagram.

aliya0001 [1]

Answer:

See the attached

Step-by-step explanation:

The highlights are attached

5 0

3 years ago

Help with my algebra homework.

just olya [345]

Q1. The answer is $\frac{(x-4)(x-4)}{(x+3)(x+1)}= \frac{ x^{2}-4x-4x+16}{ x^{2} +x+3x+3} = \frac{ x^{2} -8x+16}{ x^{2} +4x+3}$
$\frac{ x^{2} -16}{ x^{2} +5x+6} / \frac{ x^{2} +5x+4}{ x^{2} -2x-8} = \frac{ x^{2} -16}{ x^{2} +5x+6}* \frac{x^{2} -2x-8}{ x^{2} +5x+4}$
Now, factorise the numerators and denominators:
x² - 16 = x² - 4² = (x + 4)(x - 4)
x² + 5x + 6 = x² + 2x + 3x + 2*3 = x(x+2) + 3(x+2) = (x + 2)(x + 3)
x² - 2x - 8 = x² + 2x - 4x - 2*4 = x(x+2) - 4(x+2) = (x + 2)(x - 4)
x² + 5x + 4 = x² + x + 4x + 4*1 = x(x+1) + 4(x+1) = (x + 1)(x + 4)

$\frac{ x^{2} -16}{ x^{2} +5x+6}* \frac{x^{2} -2x-8}{ x^{2} +5x+4}= \frac{(x+4)(x-4)}{(x+2)(x+3)} * \frac{(x+2)(x-4)}{(x+1)(x+4)}$
Now, cancel out some factors:
$\frac{(x+4)(x-4)}{(x+2)(x+3)} * \frac{(x+2)(x-4)}{(x+1)(x+4)}= \frac{(x-4)(x-4)}{(x+3)(x+1)}= \frac{ x^{2}-4x-4x+16}{ x^{2} +x+3x+3} = \frac{ x^{2} -8x+16}{ x^{2} +4x+3}$

Q2. The answer is $\frac{7(a-7)}{(a-8)(a+8)}$
Since a² - b² = (a-b)(a+b), then a²- 64 = a² - 8² = (a-8)(a+8).
$\frac{7}{a+8} + \frac{7}{ a^{2} -64} = \frac{7}{a+8} + \frac{7}{ (a+8)(a-8)}= \frac{7(a-8)}{(a+8)(a-8)} + \frac{7}{ (a+8)(a-8)}= \frac{7(a-8)+7}{ (a+8)(a-8)}$
$= \frac{7(a-8)+7*1}{(a+8)(a-8)} =\frac{7(a-8+1)}{(a+8)(a-8)} =\frac{7(a-7)}{(a+8)(a-8)}$

Q3. The answer is $\frac{7(3a-4)}{(a-6)(a+8)}$
$\frac{ a^{2} -2a-3}{ a^{2}-9a+18 }- \frac{a^{2} -5a-6}{ a^{2}+9a+8 } = \frac{a^{2}+a-3a-3*1}{a^{2}-3a-6a+3*6} - \frac{a^{2}-a-6a-6*1}{a^{2}+a+8a+8*1}$
$= \frac{a(a+1)-3(a+1)}{a(a-3)-6(a-3)}- \frac{a(a+1)-6(a+1)}{a(a+1)+8(a+1)}= \frac{(a+1)(a-3)}{(a-6)(a-3)} - \frac{(a+1)(a-6)}{(a+1)(a+8)}$
Now, cancel out some factors:
$\frac{(a+1)(a-3)}{(a-6)(a-3)} - \frac{(a+1)(a-6)}{(a+1)(a+8)}= \frac{a+1}{a-6} - \frac{a-6}{a+8}$
$\frac{a+1}{a-6} - \frac{a-6}{a+8}= \frac{(a+1)(a+8)}{(a-6)(a+8)} -\frac{(a-6)(a-6)}{(a-6)(a+8)} =\frac{(a+1)(a+8)-(a-6)(a-6)}{(a-6)(a+8)}$
$= \frac{ a^{2} +9a+8- a^{2} +12-36}{(a-6)(a+8)} =\frac{9a+8+12-36}{(a-6)(a+8)} =\frac{21a-28}{(a-6)(a+8)} =\frac{7(3a-4)}{(a-6)(a+8)}$

Q4. The answer is $\frac{4x}{(x+3)(1+3x)}=\frac{4x}{ x^{2} +10x+3}$
$\frac{4}{x+3} / (\frac{1}{x}+3 )=\frac{4}{x+3} / (\frac{1}{x}+ \frac{3x}{x})=\frac{4}{x+3} / (\frac{1+3x}{x})= \frac{4}{x+3} * \frac{x}{1+3x} = \frac{4x}{(x+3)(1+3x)}$
$\frac{4x}{(x+3)(1+3x)}= \frac{4x}{x+3 x^{2} +3+9x}= \frac{4x}{ x^{2} +10x+3}$

Q5. The answer is x = 6
$\frac{-2}{x} +4= \frac{4}{x} +3$
$4-3= \frac{4}{x}- \frac{-2}{x}$
$1 = \frac{4-(-2)}{x}$
$1= \frac{4+2}{x}$
$1= \frac{6}{x}$
$x = 6$
Let's check the solution:
Since: $\frac{-2}{x} +4= \frac{4}{x} +3$
Then: $\frac{-2}{6}+4 = \frac{4}{6} +3$
$- \frac{1}{3}+ \frac{4*3}{3}= \frac{2}{3} + \frac{3*3}{3}$
$- \frac{1}{3} + \frac{12}{3} = \frac{2}{3} + \frac{9}{3}$
$\frac{-1+12}{3} = \frac{2+9}{3}$
$\frac{11}{3} = \frac{11}{3}$
Thus, the solution is correct

6 0

3 years ago