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mihalych1998 [28]
2 years ago
9

Please help me with my homework.

Mathematics
2 answers:
tresset_1 [31]2 years ago
8 0

Step-by-step explanation:

is the same as

x^5 + 3^5\\x^5 + 243

Does that make sense

please mark brainest

elena55 [62]2 years ago
7 0

Given expression :

{\qquad   \sf    \dashrightarrow{ (x+3)^{5} } }

We need to simplify it.

We can write the expression as,

{\qquad   \sf    \dashrightarrow{ (x+3)^{} }{ (x+3)^{} }{ (x+3)^{} }{ (x+3)^{} }{ (x+3)^{} }}

{\qquad   \sf    \dashrightarrow{ (x+3)^{2} }{ (x+3)^{2} }{ (x+3)^{} }}

We know that,

{\qquad   \sf    \dashrightarrow{ (a+b)^{2} } =  {a}^{2}  + 2ab +  {b}^{2} }

Now, using the formula in the expression:

{\qquad   \sf    \dashrightarrow{ (x {}^{2} + 2x.3 +3^{2}) }{ (x {}^{2} + 2x.3 +3^{2}) }{ (x+3)^{} }}

{\qquad   \sf    \dashrightarrow{ (x {}^{2} + 6x +9) }{ (x {}^{2} + 6x +9) }{ (x+3)^{} }}

{\qquad   \sf    \dashrightarrow{ ( {x}^{4}  + 6 {x}^{3}  + 9 {x }^{2} + 6x {}^{3}  + 36 {x}^{2} + 54x + 9 {x}^{2}   + 54x + 81 ) }{ (x+3)^{} }}

{\qquad   \sf    \dashrightarrow{ ( {x}^{4}  + 12{x}^{3}  + 54 {x }^{2} + 108x + 81 ) }{ (x+3)^{} }}

{\qquad   \sf    \dashrightarrow{ \:  {x}^{5} + 12 {x}^{4}   + 54 {x}^{3} + 108 {x}^{2} + 81x + 3 {x}^{4}   + 36x {}^{3}  + 162 {x}^{2}  + 324x + 243  }}

Adding the like terms we get :

{\qquad   \sf    \dashrightarrow{ \:  {x}^{5} + 15 {x}^{4}   + 90 {x}^{3} + 270 {x}^{2} + 405x + 243 }}

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Write as ratio in simplest form<br><br> 91%, 12%, 6%, 93%, 62%, 14%, 70%, 400%,.
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A)
91/100
we cannot simplify this

b) 12/100 = 6/50
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c) 6/100 = 3/50

d) 93/100
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e) 62/100 = 31/50

f) 14/100 = 7/50

g) 70/100 = 7/10

h) 400/100 = 4

when something is in percentage that means it is some part of 100% that is why we divide by 100.
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The mean June midday temperature in Desertville is 36°C and the standard deviation is 3°C.Assuming this data is normally distrib
AleksandrR [38]

Answer:

The value is  E(X) = 4 \ days

Step-by-step explanation:

From the question we are told that

   The mean is  \mu  =  36^oC

    The standard deviation \sigma =  3^oC

Generally the probability that in June , the midday  temperature is between  

39°C and 42°C is mathematically represented as

      P(39 <  X <  42) = P(\frac{39 - 36}{3}  < \frac{X - \mu }{\sigma} < (\frac{42 - 36}{3} )

\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )

      P(39 <  X <  42) = P(1  < Z

=>   P(39 <  X <  42) = P(Z   < 2) - P( Z

From the z table  the area under the normal curve to the left corresponding to  1 and  2  is

     P(Z   < 2)  = 0.97725

and

    P(Z   < 1)  = 0.84134

    P(39 <  X <  42) = 0.97725  - 0.84134

=>  P(39 <  X <  42) = 0.13591

Generally number of days in June would you expect the midday temperature to be between 39°C and 42°C

      E(X) =  n  *  P(39 <  X 42 )

Here n is the number of days in June which is  n = 30

       E(X) =  30  *   0.13591

=>    E(X) = 4 \ days

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