Question 1: What angle is R?
Question 2: What angle is G?

B is correct. B is rights because at the end of the equation you need to divide by 1 so it all equals out.

Answer:

Choose Choice B

Step-by-step explanation:

45-19 is 26.

All of the other equations don't make sense except B

Answer: 10 is the greatest and -8 is the least

Step-by-step explanation:

10 is the biggest number and even though -8 is a big number when you have it in the "-" the bigger the number is the less they are

start by adding the all the similar number with teh exponents but do the negatives last then you need to combine like terms and yu should get 56 as your asnwer

Answer:

48 = {1,2,3,6,8,48.}

Step-by-step explanation:

These are the factors if 48. if it's correct please let me know.

Answer:

[tex]\boxed{\mathtt{Volume=33.55m^{3}}}[/tex]

Step-by-step explanation:

[tex]\textsf{For this problem, we are asked to find the Volume of a Triangular Prism.}[/tex]

[tex]\large\underline{\textsf{What is Volume?}}[/tex]

[tex]\textsf{Volume is the amount of space a figure occupies. Volume is measured in Cubic units.}[/tex]

[tex]\large\underline{\textsf{We are Given;}}[/tex]

[tex]\textsf{The Height is 6.1m. Keep in mind that the Height is the Perpendicular Distance}[/tex]

[tex]\textsf{between 2 bases.}[/tex]

[tex]\textsf{The base is; (4.4 x 2.5).}[/tex]

[tex]\textsf{The Height of the Triangle is 2.5m.}[/tex]

[tex]\large\underline{\textsf{Necessary Formulas;}}[/tex]

[tex]\mathtt{Volume=(Base) \times Height}[/tex]

[tex]\mathtt{Area = \frac{1}{2} (Base \times Height)}[/tex]

[tex]\large\underline{\textsf{For this Problem;}}[/tex]

[tex]\mathtt{Volume=(Triangles' \ Area) \times Height}[/tex]

[tex]\large\underline{\textsf{The area of a Triangle is;}}[/tex]

[tex]\mathtt{Area = \frac{1}{2} (Base \times Height)}[/tex]

[tex]\textsf{Find the area of the base first.}[/tex]

[tex]\large\underline{\textsf{Substitute;}}[/tex]

[tex]\mathtt{Area = \frac{1}{2} (4.4 \times 2.5)}[/tex]

[tex]\boxed{\mathtt{Area = 5.5m^{2}.}}[/tex]

[tex]\textsf{We know the Base, now let's find the volume of the figure.}[/tex]

[tex]\large\underline{\textsf{Substitute;}}[/tex]

[tex]\mathtt{Volume=5.5 \times 6.1}[/tex]

[tex]\boxed{\mathtt{Volume=33.55m^{3}}}[/tex]

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textsf{Find the \boxed{\bf volume} of a triangular prism is:}[/tex]

[tex]\mathsf{\dfrac{1}{2}(\boxed{\bf b}ase\times\boxed{\bf h}eight\times\boxed{\bf l}ength) = \boxed{\bf v}olume}[/tex]

[tex]\huge\textsf{Your \boxed{\bf equation} should look like:}[/tex]

[tex]\mathsf{volume = \dfrac{1}{2}(4.4\ m \times 2.5\ m\times 6.1\ m)}[/tex]

[tex]\huge\textsf{\boxed{\bf Step-by-step} explanation can be:}[/tex]

[tex]\mathsf{volume = \dfrac{1}{2}(4.4\ m \times 2.5\ m\times 6.1\ m)}[/tex]

[tex]\mathsf{volume = \dfrac{1}{2}\times 4.4\ m \times 2.5\ m\times 6.1\ m}[/tex]

[tex]\mathsf{volume = 0.5\times 4.4\ m \times 2.5\ m\times 6.1\ m}[/tex]

[tex]\mathsf{volume = 2.2\ m\times 15.25\ m}[/tex]

[tex]\mathsf{volume = 33.55}[/tex]

[tex]\huge\textsf{Therefore your \boxed{\bf answer} should be:}[/tex]

[tex]\huge\boxed{\mathsf{volume = 33.55m^2}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

Answer:1. The measure of ∠WOV is 60°. You would use complementary angles that are adjacent (∠WOV, and ∠XOW)

2. The measure of ∠YOZ is 60°. You would use the vertical angles that are non-adjacent (∠WOV, and ∠YOZ). These two angles are congruent so they would have the same measure. These angles combined also create supplementary angles

3. Another way to find the measure of ∠YOZ would be to make/write an equation and solve for x. For example, (3x+30)°=60°. x would equal 10 because 10x3=30+30=60°

Step-by-step explanation:

1. Since a complementary angle would equal 90°, simply subtract 30° from 90° resulting in 60°.

2. Because vertical angles are congruent and (∠WOV, and ∠YOZ) are a pair of them, they equal the same as each other so they're both 60°.

3. You can make any equation with x included as long as it equals 60° mine was just an example you can make your own like 10x+10=60 or 4x+20=60. Also to create your equation you also need to use the angle fact of the vertical angles

Answer:

-6 ; 3

Step-by-step explanation:

Answer:

636.4 cm³

Step-by-step explanation:

Answer: Volume ≈ 56.55

Step-by-step explanation:

56.54867 is the answer, but I rounded it. I hope this helped! :)

Answer:

18π cubic inches

Step-by-step explanation:

First, we need to find the radius of the cone, which is half of the diameter.

radius = diameter/2 = 6/2 = 3 inches

The formula for the volume of a cone is:

V = 1/3 × π × r^2 × h

where r is the radius of the base of the cone, and h is the height of the cone.

Substituting the values we get:

V = 1/3 × π × 3^2 × 6

V = 1/3 × π × 9 × 6

V = π × 18

Answer: hope this helps ♡

1.) x = -21

2.) x = 18

3.) x = 3

Step-by-step explanation:

(minus&minus = plus)       (minus&plus = minus)        (plus&minus = minus)

1.) 25 - 3x = 88

25 - 25 - 3x = 88 - 25

-3x = 63

-3x ÷ 3 = 63 ÷ 3

-x = 21 → x = -21

2.) -11 = 7 - x

-11 - 7 = 7 -7 -x

-18 = -x

18 = x → x = 18

3.) 95 = -4 + 33x​

95 + 4 = -4 + 4 + 33x

99 = 33x

99 ÷ 33 = 33x ÷ 33

3 = x → x = 3

[tex] \text{1 ] 25 - 3x = 88 }[/tex]

[tex] \rm{↣25 - 3x = 88} \\ \rm↣- 3x= 88 - 25 \\ \rm↣- 3x = 63 \\ \rm↣x = \cancel \frac{63}{ - 3} \\ \underline{ \underline{\rm ↣\purple{ \: x = -21 \: }}}[/tex]

[tex] \: [/tex]

_________

[tex] \text{2 ] -11 = 7 - x }[/tex]

[tex] \rm \: ↣-11 = 7 - x \\ \rm↣ -11 - 7 = -x \\ \rm↣ \cancel- 18 = \cancel -x\\\underline{ \underline{ \rm↣ \pink{x = 18}}}[/tex]

[tex] \: [/tex]

_________

[tex] \text { 3 ] 95 = -4 + 33x }[/tex]

[tex] \rm \: ↣ 95 = -4 + 33x \\ \rm↣ 95 + 4 = 33x \\ \rm↣ 99 = 33x \\ \rm ↣ \cancel \frac{99}{33} = x \\ \rm ↣3 = x \\ \underline{ \underline{\rm \color{hotpink}↣x = 3}}[/tex]

[tex] \: [/tex]

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hope it helps:)

Step-by-step explanation:

A triangle has its interior angle sum as 180°

[tex]{ \tt{ \angle R + \angle S + \angle T = 180 \degree}}[/tex]

[tex]{ \colorbox{silver}{input \: angles}} \\ \\ { \tt{31 \degree + (x + 4) \degree + (3x + 9) \degree = 180 \degree}} \\ \\ { \tt{44 + 4x = 180 \degree}} \\ \\ { \tt{4x = 136}} \\ \\ { \tt{x = 34}} \\ \\ { \colorbox{silver}{for \:R }} \\ \\ { \tt{ m\angle S =x + 4 = 34 + 4}} \\ \\ { \boxed{ \tt{m \angle S = 38 \degree}}} \\ \\ { \colorbox{silver}{for \:T }} \\ \\ { \tt{m \angle T = 3x + 9 = (3 \times 34) + 9}} \\ \\ { \tt{m \angle T = 111 \degree}}[/tex]