Can anyone solve this?

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:

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:

636.4 cm³

Step-by-step explanation:

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]