c. Which volume-to-surface-area ratio would be better for an ice cube¡X the lowest possible or the highest possible?

Final answer:

A triangle with two perpendicular sides could be scalene or isosceles if the non-perpendicular sides are unequal or equal, respectively. It cannot be equilateral because equilateral triangles have no right angles.

Explanation:

If a triangle has two sides that are perpendicular to each other, it could potentially be a scalene or isosceles triangle, but it cannot be equilateral. When two sides of a triangle are perpendicular, they form a right angle. Thus, if a triangle has a right angle, it is a right triangle by definition. According to Theorem 2, if two angles of a triangle are equal, the opposite sides are equal, and the triangle is isosceles.

In our case, this theorem would apply if the two non-perpendicular sides are of equal length, resulting in an isosceles right triangle. On the other hand, Theorem 3 states that in a triangle with unequal angles, the side opposite the greater of the angles is greater than the side opposite the smaller. Therefore, if the sides forming the right angle are of different lengths, the triangle would be scalene.

Given is a Right triangle LMN with right angle at M i.e. angle M = 90 degrees.

Given are the sides LM = 15, MN = 36, and LN = 39.

It says to find cos(N) = ?

So opposite side would be LM, adjacent side would be MN, and LN is the hypotenuse.

We know about the "Soh-Cah-Toa" rule in which cosine ratio is given by :-

[tex] cos(N) = \frac{Adjacent}{Hypotenuse} \\\\
cos(N) = \frac{MN}{LN} \\\\
cos(N) = \frac{36}{39} \\\\
cos(N) = \frac{12}{13} [/tex]

Hence, final answer is [tex] cos(N) = \frac{12}{13} [/tex].

To find the surface area of a cylinder sandwiched between one cone and half of a sphere, use the sum of the cylinder's lateral surface area, the cone's lateral surface area, and half the sphere's surface area. The formula is 2πrh + πrl + 2πr², excluding the base areas where the shapes intersect.

To determine the surface area of a cylinder sandwiched between one cone and half of a sphere, you should first consider the formulas for the surface area of these individual shapes. The shapes consist of simple geometrical forms: a cylinder, a cone, and a sphere. The surface area of these shapes can be described with established formulas within the geometry:

Therefore, the final formula to calculate the total surface area of the described complex shape is the sum of the surface area of the lateral side of the cylinder, the lateral surface area of the cone, and half of the surface area of the sphere. This could be written as:
Total Surface Area = 2πrh + πrl + 2πr².

Note that you would omit the area of the circle that is common between the cylinder and the cone, as well as the area that is common between the cylinder and the half-sphere, since they are internall and not part of the external surface area.

The length of QR in rectangle PQRS is 12 cm, found by setting up the proportional relationship 14/21 = 8/QR, where 14 cm and 8 cm are the lengths of sides AB and BC of rectangle ABCD, respectively, and 21 cm is the length of side PQ of rectangle PQRS.

Since rectangle ABCD is similar to rectangle PQRS, their corresponding sides are proportional. To find the length of QR, we will set up a proportion using the given lengths:

AB/PQ = BC/QR

Substituting the given lengths:

14/21 = 8/QR

Now, solve for QR:

QR = (21  * 8) / 14

QR = 168 / 14

QR = 12 cm

So, the length of QR in rectangle PQRS is 12 cm.

Final answer:

The first step of rationalizing the denominator of 7√11 is to multiply both the numerator and denominator by the conjugate of the denominator.

Explanation:

The first step of rationalizing the denominator of 7√11 is to multiply both the numerator and the denominator by the conjugate of the denominator. In this case, the conjugate of √11 is -√11. Therefore, we multiply both the numerator and denominator by -√11 to get:

7√11 * -√11 = -77

√11 * -√11 = -11

So, the rationalized denominator is -11. The denominator becomes -11, and the simplified expression is:

-77 / -11 = 7

Answer:

Option a.

Step-by-step explanation:

Two rectangles are similar. Height of one rectangle is 3 inches and height of second rectangle is 9 inches.

Let width of rectangles are x inches and y inches.

Then width of the second rectangle will be in the same ratio as of their heights.

[tex]\frac{x}{y} =\frac{3}{9}[/tex]

[tex]\frac{x}{y} =\frac{1}{3}[/tex] ⇒ x = [tex]\frac{y}{3}[/tex]

Now perimeter of first rectangle P₁ = 3 + 3 + x + x  = 2x + 6

Perimeter of second rectangle P₂ = 9 + 9 + y + y = 18 + 2y

Ratio of P₁ and P₂ = [tex]\frac{2x+6}{18+2y}[/tex]

                             = [tex]\frac{2(\frac{y}{3})+6}{18+2y}[/tex]   [as [tex]x=\frac{y}{3}[/tex]]

                            = [tex]\frac{\frac{(2y+18)}{3} }{(18+2y)}[/tex]

                            = [tex]\frac{2y+18}{3(18+2y)}[/tex]

Ratio of P₁ and P₂ = ([tex]\frac{1}{3}[/tex]) ⇒ P₂ = 3P₁

Therefore, Option a is the answer.

Final answer:

The number 4,321,109,432 rounded to the nearest ten million is 4,320,000,000,hence the correct option is B.

Explanation:

To round the number 4,321,109,432 to the nearest ten million, we need to look at the digit in the ten million's place and the digit to the right of it, which are 3 and 1, respectively. The rule for rounding is that if the digit to the right is 5 or greater, we round up.

In this case, the digit is 1, so we do not round up and the ten million's place remains unchanged.

Therefore, the answer is 4,320,000,000 (Option B), as every digit after the ten million's place is replaced with zero.

In this geometric scenario, angles are named and classified. Angle 1 is a right angle, angle 3 is a straight angle, angles 2 and 6 are obtuse, while angles 4 and 5 are acute.

Let's name each angle in four ways and classify them:

Angle ABC (B is 90 degrees):

1 (as labeled)

∠ABC (using the vertex B)

∠CBA (using the vertex C)

∠1 (angle 1)

Classification: Right angle

Angle DEF (E is also labeled as 2):

2 (as labeled)

∠DEF (using the vertex E)

∠FED (using the vertex F)

∠2 (angle 2)

Classification: Obtuse angle

On line LN with point M (labeled as 3):

3 (as labeled)

∠LNM (using the vertex N)

∠MNL (using the vertex M)

∠3 (angle 3)

Classification: Straight angle

Angle XYZ (Y is also labeled as 4):

4 (as labeled)

∠XYZ (using the vertex Y)

∠ZYX (using the vertex Z)

∠4 (angle 4)

Classification: Acute angle

Angle KLM (L is also labeled as 5):

5 (as labeled)

∠KLM (using the vertex L)

∠MLK (using the vertex M)

∠5 (angle 5)

Classification: Acute angle

Angle RSP (S is also labeled as 6):

6 (as labeled)

∠RSP (using the vertex S)

∠PSR (using the vertex P)

∠6 (angle 6)

Classification: Obtuse angle

So, to summarize:

Right angle: 1

Obtuse angles: 2, 6

Straight angle: 3

Acute angles: 4, 5

To know more about angle:

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