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1 year ago

A cone without a base is made from a half-circle of radius 10 cm. Determine the volume of the cone. Explain your reasoning.

Mathematics

1 answer:

DochEvi [55]1 year ago

5 0

Answer:

125π√3/3 cm³ ≈ 226.72 cm³

Step-by-step explanation:

The length of the circular edge of the half-circle is ...

(1/2)C = (1/2)(2πr) = πr = 10π . . . . cm

This is the circumference of the circular edge of the cone, so the radius of the cone is found from ...

C = 2πr

10π = 2πr . . . . fill in the numbers; next, solve for r

r = 5 . . . . cm

The slant height of the cone is the original radius, 10 cm, so the height of the cone from base to apex is found from the Pythagorean theorem.

(10 cm)² = h² + r²

h = √((10 cm)² -(5 cm)²) = 5√3 cm

And the cone's volume is ...

V = 1/3·πr²h = (1/3)π(5 cm)²(5√3 cm)

V = 125π√3/3 cm³ ≈ 226.72 cm³

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Which expression is equivalent to 7a2b + 10a2b2 + 14a2b3?

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The given expression can be simplified in many ways by grouping like terms. The simplest form is obtained by factoring out a²b which gives us the following expression.

a²b(7 + 10b +14b²)

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2 years ago

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Write the fraction in ascending order.

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from most to least:

i: 1/2, 1/3, 1/7, 1/10

ii: 2/3, 2/5, 2/9, 2/11

iii: 5/6, 3/4, 2/3, 1/2

iv: 4/5, 3/4, 5/8, 1/2

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1 year ago

Tara owes $14,375 in credit card debt. The interest accrues at a rate of 5.3%. She is also borrowing $570 each month for rent fr

siniylev [52]

Answer:

$(f + g)(t) = f(t) + g(t) = 14375 (1 + \frac{5.3}{100})^{t} + 6840t$

$29619.13

Step-by-step explanation:

a. Tara has $14375 in credit card debt and the interest rate is 5.3%.

Now, if f(t) represent the amount of money Tara have in credit card debt, where t is the number of years after after interest begins to accrue, then

$f(t) = 14375 (1 + \frac{5.3}{100})^{t}$ ......... (1)

Again Tara borrows $570 each month for rent from her parents without any interest.

If g(x) represent the amount of money Tara owes to her parents, where t represents the number of years passed,then we can write

g(t) = 570 × 12t = 6840t ........ (2)

Therefore, $(f + g)(t) = f(t) + g(t) = 14375 (1 + \frac{5.3}{100})^{t} + 6840t$

b. So, for t = 2 years,

$(f + g)(t) = 14375 (1 + \frac{5.3}{100})^{2} + 6840 \times 2$ = $29619.13

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1 year ago

Evaluate 8a-1+0.5b8a−1+0.5b8, a, minus, 1, plus, 0, point, 5, b when a=\dfrac14a=

professor190 [17]

Answer: $6$

Step-by-step explanation:

Given the following expression:

$8a-1+0.5b$

You need to substitute the given values of "a" and "b" into the expression. Notice that these values are:

$a=\frac{1}{4}\\\\b=10$

Then;

$8a-1+0.5b=8(\frac{1}{4})-1+0.5(10)$

Now you must solve the multiplications:

$=\frac{8}{4}-1+5=2-1+5$

The final step is to add the numbers. Therefore, you get the following answer:

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2 years ago

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The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si

natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b) y = 1.148*x + 6.036

Step-by-step explanation:

Given:

f(x) = 6 - 10*x^2

g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

f'(x) = -20*x

g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

m = (g(x_o) - f(x)) / (x_o - x)

m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

m = f'(x) = g'(x_o)

- We will develop the first expression:

m = f'(x)

( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1. (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

m = g'(x_o)

( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

-2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2 -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

-20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

-22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors: 20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

(x - x_o)(20*x - 2*x_o + 4) = 0

x = x_o , x_o = 10x + 2

- For x_o = 10x + 2 ,

(g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

(8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

(-90*x^2 + 2) = -180*x^2 - 40*x

90*x^2 + 40*x + 2 = 0

- Solve the quadratic equation above:

x = -0.0574, -0.387

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y - 4.502 = -20*(-0.387)*(x + 0.387)

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- Other tangent line:

y - 5.97 = 1.148*(x + 0.0574)

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6 0

1 year ago