If a = 3 and b = 5, find a/b

Circumference, radius, slant height, lateral area, and surface area are measurements of a cone. Given two of the measurements, f

USPshnik [31]

Answer:

LA = 21.7 ft²; C = 18.8 ft; h = 2.3 ft
h = 8 in; r = 7 in; SA = 330.1 in²

Step-by-step explanation:

Applicable relations are ...

C = 2πr . . . . . . . r = radius, C = circumference

LA = 1/2·Ch . . . . h = slant height, LA = lateral area

SA = πr² +LA . . . SA = total surface area

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1. Given r and SA, we can find the other measures as ...

LA = SA -πr² = 50 ft² -π(3 ft)² ≈ 21.7 ft²

C = 2πr = 2π(3 ft) ≈ 18.8 ft

h = 2·LA/C ≈ 2·21.7 ft²/(18.8 ft) ≈ 2.3 ft

(Please note that for calculations using intermediate results, we used the full-precision values of those results, not the rounded ones. Rounding is always the last operation. Please note, too, that we have used a value for π sufficient to ensure accuracy in all the digits shown here.)

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2. Given C and LA, we can find the other measures as ...

h = 2·LA/C = 2(176 in²)/(44 in) = 8 in

r = C/(2π) = (44 in)/(2π) ≈ 7.0 in

SA = LA +πr² = LA +rC/2 ≈ 176 in² +(7.0 in)(44 in)/2 ≈ 330.1 in²

4 0

3 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

The difference of a number x and 9 is fewer then 4

Sav [38]

The equation would be:
x - 9 < 4
Simplify (add 9)
Solution: x < 13

8 0

3 years ago

Write an equation of the perpendicular bisector of the line segment whose endpoints are (−1,1) and (7,−5)

icang [17]

The equation is $y = \frac{3}{2} x - \frac{11}{2}$

<u>Explanation:</u>

We have to first find the mid-point of the segment, the formula for which is

$(\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2} )$

So, the midpoint will be $(\frac{-1+7}{2} , \frac{1-5}{2} )\\\\$

= $(3,-2)$

It is the point at which the segment will be bisected.

Since we are finding a perpendicular bisector, we must determine what slope is perpendicular to that of the existing segment. To determine the segment's slope, we use the slope formula $\frac{y_2-y_1}{x_2-x_1}$