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1 year ago

Sam paid $50 for her new skateboard. In addition, she paid $3.50 in taxes. What was the tax rate on Sam's purchase?

Mathematics

1 answer:

navik [9.2K]1 year ago

8 0

Answer: $53.50 ?

Step-by-step explanation:

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Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

Using Pythagoras Theorem

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

First derivative

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

apply chain rule

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

Equate the first derivative to zero, that is V'(x) = 0

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$

maximum volume: when h = 40/3

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

minimum volume: when h = 0

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

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1 year ago

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48÷479×390÷38789946=

ankoles [38]

correct answer is 0.0000010075

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Find the value of x. Please help

vitfil [10]

Answer:

The answer is 57.

180 - 67 - 56 = 57

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What two numbers multiply to 10 and add to -29 ?

Degger [83]

Xy=10
x+y=29

subtract x from both sides for second equation

y=-x-29
subsitute in second equaiton
x(-x-29)=10
-x^2-29x=10
add x^2+29x to both sides
x^2+29x+10=0
quadratic formula
if you have
ax^2+bx+c=0,
x= $\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}$

so
x^2+29x+10=0
a=1
b=29
c=10
x= $\frac{-29+/- \sqrt{(-29)^{2}-4(1)(10)} }{2(1)}$
x= $\frac{-29+/- \sqrt{841-40} }{2}$
x= $\frac{-29+/- \sqrt{801} }{2}$
x= $\frac{-29+/- 3\sqrt{89} }{2}$

x= $\frac{-29- 3\sqrt{89} }{2}$ or $\frac{-29+ 3\sqrt{89} }{2}$
those are the 2 numbers
aprox=-28.65 and -0.349

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54 because the formula for the area of triangle is 1/2bh so you just substitute the b and h for 18 and 6

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