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sladkih [1.3K]
3 years ago
11

Matteo spends a total of 38 min exercising. He walks for 6 min to warm up and then runs at a constant rate of 8 min per mile for

the rest of the time. Matteo says that he ran 4.75 mi. Is he correct?
Mathematics
1 answer:
umka2103 [35]3 years ago
6 0

Answer:

No, he not correct because he only ran 4 mi

Step-by-step explanation:

If he walks for 6 min out of the 38 min, then he ran for 32 min.

d = rt        1 mi = 8r

r = 1/8 = 1 mi/8 min   Since he ran 32 min we have

d = 1/8(32) = 4 mi.

No, he not correct because he only ran 4 mi

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Why 10^33/2=5*10^32? Explain, pls. :(
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Answer:

see below

Step-by-step explanation:

10^33/2

Rewriting the numerator as 10 * 10 ^ 32

10 * 10 ^32

--------------------

2

10/2 = 5

5 * 10 ^32

Therefore

10^33/2=5*10^32

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3 years ago
Sin0=12/37 find tan0
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Answer:

\large{ \tt{❃ \: EXPLANATION}} :

  • We're provide - Sin θ = \frac{12}{37} which means 12 is the perpendicular & 37 is the hypotenuse [ Since Sin θ = \tt{  \frac{p}{h}} ] . We're asked to find out tan θ ].

\large{ \tt{❁ \: USING \: PYTHAGORAS \: THEOREM}} :

\large{ \tt{❊ \:  {h}^{2}  =  {p}^{2}  +  {b}^{2} }}

\large{ \tt{⇢ {p}^{2} +  {b}^{2}   =  {h}^{2} }}

\large{ \tt{⇢ \:  {b}^{2}  =  {h}^{2}  -  {p}^{2} }}

\large{ \tt{⇢ \:  {b}^{2} = {37}^{2} -  {12}^{2}   }}

\large{ \tt{⇢ \:  {b}^{2}  = 1369 - 144}}

\large{ \tt{ ⇢{b}^{2}  = 1225}}

\large{ \tt{⇢ \: b =  \sqrt{1225}}}

\large{ \tt{⇢ \: b = 35  \: \text {units}}}

  • Now , We know - Tan θ= \tt{ \frac{perpendicular}{base} }. Just plug the values :

\large{ \tt{➝ \: Tan  \: \theta =  \frac{p}{b}  = \boxed{ \tt{  \frac{12}{35} }}}}

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

5 0
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Step-by-step explanation:

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