= [3x3^-2x(-2)^6/2x3^-1x(-2)^5]^2 = 1
Answer:
The area of the circle is 19.5 in²
Step-by-step explanation:
First of all to solve this problem we have to know the formula to calculate the volume of a cone
v = volume = 52 in³
r = radius
h = height = 8 in
π = 3.14
v = 1/3 * π * r² * h
we solve r
3 * v /h * π = r²
we replace the known values
3 * 52 in³ / 3.14 * 8 in = r²
156 in³ / 25.12 in = r²
6.21 in² = r²
√6.21 in² = r
2.49 in = r
now that we have the radius we need to use the area formula of a circle:
a = area
r = radius = 2.49 in
π = 3.14
a = π * r²
we replace the known values
a = 3.14 * (2.49 in)²
a = 3.14 * 6.21 in²
a = 19.5 in²
The area of the circle is 19.5 in²
Answer:
4 √6
Step-by-step explanation:
We have a few right triangles. We know that a²+b²=c², with c being the side opposite the right angle. Representing the side without a value as z, we have:
m²+z² = (8+4)² = 12²
4²+n²=z²
8²+n²=m²
We have 3 equations with 3 unknown variables, so this should be solvable. One way to find a solution is to put everything in terms of m and go from there. First, we can take n out of the equations entirely, removing one variable. We can do this by solving for it in terms of z and plugging that into the third equation, removing a variable as well as an equation.
4²+n²=z²
subtract 4²=16 from both sides
z²-16 = n²
plug that into the third equation
64 + z² - 16 = m²
48 + z² = m²
subtract 48 from both sides to solve for z²
z² = m² - 48
plug that into the first equation
m² + m² - 48 = 144
2m² - 48 = 144
add 48 to both sides to isolate the m² and its coefficient
192 = 2m²
divide both sides by 2 to isolate the m²
96 = m²
square root both sides to solve for m
√96 = m
we know that 96 = 16 * 6, and 16 = 4², so
m = √96 = √(4²*6) = 4 √6
Answer:
Its magnitude will be larger than 0.004.
Step-by-step explanation:
When a divisor is less than 1, the quotient will be greater than the dividend.
When the divisor is "almost zero", the quotient will be much greater than the dividend. Here, the dividend may be considered to be "almost zero", so we cannot say anything about the actual quotient except to say its magnitude will be greater than the dividend.
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The dividend is positive, so the quotient will have the same sign as the divisor. (Negative divisors can be "almost zero," too.)
