The hight of a radius of a cine ara measured as 21cm and 14.0cm respectively . Taking 3.142. find the% area in the volume of the cone .
To subtract fractions, we need to have a common denominator. In this case, the denominators are 2 and 6, which have a common multiple of 6. So we need to rewrite each fraction with a denominator of 6:
(4x - 5)/2 - (2x - 1)/6
= (3*(4x - 5))/(3*2) - (1*(2x - 1))/(1*6)
= (12x - 15)/6 - (2x - 1)/6
Now that we have a common denominator, we can subtract the two fractions:
= (12x - 15 - 2x + 1)/6
= (10x - 14)/6
We can further simplify this fraction by factoring out a 2 from the numerator:
= 2(5x - 7)/6
= (5x - 7)/3
Therefore, the single fraction equivalent of (4x - 5)/2 - (2x - 1)/6 is (5x - 7)/3.
We can use the formula for the volume of a cone:
V = (1/3)πr^2h
where r is the radius and h is the height.
The radius is given as 14.0 cm and the height as 21 cm. Substituting these values and using 3.142 as the value of π, we have:
V = (1/3)π(14.0)^2(21)
V = 10,164.856 cm^3
Now, let's consider the percentage of the volume that is occupied by the cone with the given radius and height. To do this, we need to find the volume of the full cone that has the same radius and height.
The formula for the volume of a cone is:
V = (1/3)πr^2h
In this case, the radius and height are both 21 cm. So we have:
V = (1/3)π(21)^2(21)
V = 24,003.594 cm^3
The percentage of the volume that is occupied by the cone with the given radius and height is:
(10,164.856 / 24,003.594) x 100%
= 42.34%
Therefore, the cone with the given radius and height occupies approximately 42.34% of the total volume of the cone.
Point B is 3cm away from point A of a bearing 150°c point C is 25cm from A at a bearing . Find how far is from B.
First, we need to draw a diagram to visualize the problem:
```
B
/
/
/
3cm / 150°
/ \
/ \
/ \
A C
\
\
\
\
25cm
\
\
\
\
180°c
```
From the diagram, we can see that we have a triangle ABC. We want to find the distance between B and C.
To do this, we can use the law of cosines:
c^2 = a^2 + b^2 - 2ab*cos(C)
where a, b, and c are the sides of the triangle and C is the angle opposite side c.
We know that AB = 3 cm, AC = 25 cm, and the angle between them is 150°. So we can find BC:
BC^2 = 3^2 + 25^2 - 2*3*25*cos(180° - 150°)
BC^2 = 784
BC = sqrt(784)
BC = 28 cm
Therefore, the distance between B and C is 28 cm.
Reduce the following expression on to a single fraction (4x-5/2)-(2x-1/6)
Make n the subject of the fomular . p(1+1/100)^n
We have an expression:
p(1 + 1/100)^n
To make n the subject of this expression, we need to isolate n on one side of the equation.
First, we can simplify the expression:
p(1 + 1/100)^n = p(101/100)^n
Now, we can take the natural logarithm of both sides:
ln(p(101/100)^n) = ln(p) + ln((101/100)^n)
Using the properties of logarithms, we can simplify this to:
ln(p) + n*ln(101/100) = ln(p(101/100)^n)
Next, we can isolate n by subtracting ln(p) from both sides:
n*ln(101/100) = ln(p(101/100)^n) - ln(p)
Using the quotient rule of logarithms on the right-hand side, we get:
n*ln(101/100) = ln[(p(101/100)^n)/p]
Simplifying the right-hand side:
n*ln(101/100) = ln(101/100)^n
n*ln(101/100) = n*ln(101/100)
Now, we can divide both sides by ln(101/100):
n = [ln(p(101/100)^n)/p] / ln(101/100)
Therefore, the formula for n is:
n = [ln(p(101/100)^n)/p] / ln(101/100)
Find the equation of a perpediculer bisector PQ from y=mx+c if the coordinates of P and Q are (-2,6) and 4,-2) respectively .
The perpendicular bisector of a line segment connects the midpoint of the line segment to its perpendicular bisector. This means that the perpendicular bisector of PQ will go through the midpoint of PQ, which we can find using the midpoint formula:
Midpoint = [(-2 + 4)/2, (6 + (-2))/2] = [1, 2]
So P is (-2, 6) and Q is (4, -2), and the midpoint of PQ is (1, 2).
Now, we need to find the slope of the line y = mx + c that we want to find the perpendicular bisector of. We can see that the slope of this line is m.
The slope of a line perpendicular to y = mx + c will be the negative reciprocal of m. So the slope of the perpendicular bisector, let's call it b, will be:
b = -1/m
Now, we can use the point-slope form of the equation of a line to find the equation of the perpendicular bisector. We have a point (1, 2) and a slope b. So the equation of the line is:
y - 2 = b(x - 1)
Substituting in b = -1/m, we get:
y - 2 = (-1/m)(x - 1)
Multiplying both sides by -m, we get:
-m(y - 2) = x - 1
Simplifying, we get:
my - 2m = x - 1
Rearranging, we get the final equation of the perpendicular bisector:
x - my = 2m + 1
Therefore, the equation of the perpendicular bisector PQ in terms of y = mx + c is x - my = 2m + 1.
To find the percentage of the area in the volume of a cone, we need to calculate the area of the base and the volume of the cone.
First, let's find the area of the base of the cone using the formula for the area of a circle:
Area of the base = π × radius²
Given that the radius is 14.0 cm, we can calculate the area of the base:
Area of the base = 3.142 × 14.0² cm² (rounding the value of π to 3.142)
Next, let's find the volume of the cone using the formula:
Volume = (1/3) × π × radius² × height
Given that the height is 21 cm and the radius is 14.0 cm, we can calculate the volume:
Volume = (1/3) × 3.142 × 14.0² × 21 cm³
Now, to find the percentage of the area in the volume of the cone, we divide the area of the base by the volume of the cone and multiply by 100:
Percentage of area = (Area of the base / Volume) × 100
Plug in the values we calculated earlier:
Percentage of area = (3.142 × 14.0² cm²) / [(1/3) × 3.142 × 14.0² × 21 cm³] × 100
Cancel out the common factors:
Percentage of area = (1) / [(1/3) × 21] × 100
Simplify the expression:
Percentage of area = (1) / 7 × 100
Percentage of area = 14.29%
Therefore, the percentage of the area in the volume of the cone is approximately 14.29%.