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Anon25 [30]
3 years ago
11

For basketball tryouts, Kiana ran two laps in a total of 28.5 seconds. In the second lap, she beat her first lap by 3 seconds. W

rite an equation to represent the situation. I a clase Solve the equation to calculate how long to run the second lap?​
Mathematics
1 answer:
yanalaym [24]3 years ago
8 0

Answer:add 6

Step-by-step explanation:

yo have to add 5 then subtract to get answer

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49/63 reduced fraction
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49 / 7 = 7
63 / 7 = 9,
You place 7 over 9 and you get your answer!

7/9
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Which is the decimal expansion of 7 Over 22
Fed [463]

Answer: 0.31

Step-by-step explanation:

7/22 is 0.31

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If LM = MP, then M must be the midpoint of LP? True or False? Im so confused on this!
ikadub [295]

Answer:

true

Step-by-step explanation:

if you have a line LP and midpoint M means M divides LP in to two equal parts ie LM and MP

6 0
2 years ago
The length of the sides DEF are 12,16,and 20. Find the perimeter of the triangle formed by connection the midpoints of the sides
Mekhanik [1.2K]

Answer:

  • 24 units

Step-by-step explanation:

According to the midsegment theorem, the length of each side of the formed triangle is the half the parallel side of DEF.

<u>So the perimeter is:</u>

  • P = 1/2(12 + 16 + 20) = 24
6 0
2 years ago
12. In the given figure, RS is parallel to PQ, If RS = 3 cm, PQ = 6 cm and ar(∆TRS) = 15cm³, then ar (∆TPQ) = ? (a) 70 cm² (b) 5
Gnesinka [82]

\large\underline{\sf{Solution-}}

Given that,

In <u>triangle TPQ, </u>

  • RS || PQ,

  • RS = 3 cm,

  • PQ = 6 cm,

  • ar(∆ TRS) = 15 sq. cm

As it is given that, <u>RS || PQ</u>

So, it means

⇛∠TRS = ∠TPQ [ Corresponding angles ]

⇛ ∠TSR = ∠TPQ [ Corresponding angles ]

\rm\implies \: \triangle TPQ \:  \sim \: \triangle TRS \:  \:  \:  \:  \:  \:  \{AA \}

<u>Now, We know </u>

Area Ratio Theorem,

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{ar( \triangle \: TRS)}  = \dfrac{ {PQ}^{2} }{ {RS}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{ {6}^{2} }{ {3}^{2} }

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = \dfrac{36 }{9}

\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15}  = 4

\rm\implies \:ar( \triangle \: TPQ)  = 60 \:  {cm}^{2}

3 0
2 years ago
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