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qaws [65]
2 years ago
5

70 points!! A conjecture and the flowchart proof used to prove the conjecture are shown.

Mathematics
2 answers:
ikadub [295]2 years ago
3 0

Answer:

top left, S is the midpoint of RT, given

left 2nd row, US=RS is the definition of midpoint

right 2nd row, US=RS

3rd row,US is congruent to ST, definition of an isosceles triangle

4th row, triangle STU is an isosceles triangle, due to SSS

Hope this helps! :)

luda_lava [24]2 years ago
3 0

Answer:

um where are the answer choices, and did you ever get the answer to this?

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Enguun earns $17 per hour tutoring student-athletes at Brooklyn University.
Juliette [100K]

Answer:

a. Earnings of Enguun for 12 hours this month (G)

G = 204 dollars

b. Earnings of Enguun for  19.5 hours last month (G)

G = 331.5 dollars

Step-by-step explanation:

Nomenclature

G: Total earnings of Enguun in dollars

E: Enguun's earnings for 1 hour of tutoring (dollars/hour)

n: number of hours that Enguun tutored per month (hour/month)

Formula

G = E*n

E = $17 per hour = 17 dollars/hour

Problem development

a. Earnings of Enguun for 12 hours this month

n = 12 hours

G = E*n

G= 17\frac{dollars}{hour} * 12 hour  : We eliminate hours

G = 204 dollars

b. Earnings of Enguun for 19.5 hours last month

G = E*n

G = 17 * 19.5

G = 331.5 dollars

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3 years ago
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Can someone help with a really simple median question? tbh i really don't know it :(
Goryan [66]
I believe the answer is the second answer
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3 years ago
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Consider the figure. What is the area of the figure, in square centimeters? ​
Serhud [2]

Answer: The

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2 years ago
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The ammunition storage room has 10 feet between the floor and the ceiling. Each box of ammunition is 10 feet tall. Each crate of
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Answer:

the maximum number of crates that can be stacked between the floor and ceiling

\frac{height \hspace{0.1cm} of  \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of  \hspace{0.1cm} each \hspace{0.1cm} crate}  = \frac{10 \hspace{0.1cm}  feet}{12 \hspace{0.1cm}  inches}  =  \frac{10 \times 12  \hspace{0.1cm} inches }{12  \hspace{0.1cm} inches}  = 10 \hspace{0.1cm}  crates

where 1 foot  = 12 inches. SO the answer is that a maximum of 10 crates can be stacked from floor to ceiling.

Step-by-step explanation:

i) the maximum number of crates that can be stacked between the floor and ceiling

\frac{height \hspace{0.1cm} of  \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of  \hspace{0.1cm} each \hspace{0.1cm} crate}  = \frac{10 \hspace{0.1cm}  feet}{12 \hspace{0.1cm}  inches}  =  \frac{10 \times 12  \hspace{0.1cm} inches }{12  \hspace{0.1cm} inches}  = 10 \hspace{0.1cm}  crates

where 1 foot  = 12 inches

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2 years ago
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