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Dafna11 [192]
3 years ago
6

Emma needs to earn at least $250 a week during her summer break to pay for college. She works two jobs, one at a gas station tha

t pays $25 an hour and the other delivering groceries for $14.50 per hour.
Let x be the number of hours she works at the gas station and let y be the number of hours she works delivering groceries. Write an inequality that models this situation.
Mathematics
1 answer:
ANTONII [103]3 years ago
4 0

Answer:

x×25+y×14.5( greater than or equal to symbol)250

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Harold uses the binomial theorem to expand the binomial (3x^5 - 1/9y^3)^4
riadik2000 [5.3K]
<h3><em>The complete question:</em></h3>

<u><em> </em></u><u>Harold uses the binomial theorem to expand the binomial </u>(3x^5 -\dfrac{1}{9}y^3)^4<u />

<u>(a)    What is the sum in summation notation that he uses to express the expansion? </u>

<u>(b)    Write the simplified terms of the expansion.</u>

Answer:

(a). (3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n}  \binom{4}{k}(3x^5)^{4-k}( -\dfrac{1}{9}y^3)^k $$

(b).(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}

Step-by-step explanation:

(a).

The binomial theorem says

(x+y)^n=$$\sum_{k=0}^{n}  \binom{n}{k}x^{n-k}y^k $$

For our binomial this gives

\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n}  \binom{4}{k}x^{4-k}y^k $$}

(b).

We simplify the terms of the expansion and get:

$$\sum_{k=0}^{n}  \binom{4}{k}(3x^5)^{4-k}y^k $$= \binom{4}{0}(3x^5)^{4-0}(-\dfrac{1}{9}y^3 )^0+\binom{4}{1}(3x^5)^{4-1}(-\dfrac{1}{9}y^3 )^1+\\\\\binom{4}{2}(3x^5)^{4-2}(-\dfrac{1}{9}y^3 )^2+\binom{4}{3}(3x^5)^{4-3}(-\dfrac{1}{9}y^3 )^3+\binom{4}{4}(3x^5)^{4-4}(-\dfrac{1}{9}y^3 )^4

$$\sum_{k=0}^{n}  \binom{4}{k}(3x^5)^{4-k}(-\frac{1}{9}y^3 )^k $$= (3x^5)^{4}+4(3x^5)^{3}(-\frac{1}{9}y^3 )+6(3x^5)^{2}(-\frac{1}{9}y^3 )^2+\\\\4(3x^5)(-\frac{1}{9}y^3 )^3+(-\frac{1}{9}y^3 )^4

\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}   }

3 0
3 years ago
PLZ HELP ill GIVE BRAINIESt!!!!20 Points
Ronch [10]
A) √50 = √(25 x 2) =√(5² x 2) = 5√2, what Jacklyn did is the other way round, instead of putting the perfect square out of the radical & keep the 2 inside, she inversed the sens of the operation

b) We have to find the square of the smaller & largest numbers that are near 50:

7² = 49   & 8² = 64==> so 49<50<64 & the number is between the square root of 7 & the square root of 8, but we also notice that 50 is very very near 49, hence let's try 7.1==> 7² = 7.1 x 7.1 = 50.41, which is a very good approximation. Then the approx. to √50 ≈ 50.41
7 0
4 years ago
Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +
Veronika [31]

Answer:

Verified

y(x) = \frac{3Ln(x) + 3}{x}

y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

Download docx
6 0
3 years ago
What’s the answer to this
lana66690 [7]

250-number\ of\ people\\\\\dfrac{2}{5}\ of\ the\ 250\ people\to\dfrac{2}{5}\cdot250=\dfrac{2}{1}\cdot50=100-number\ of\ girls\\\\\dfrac{3}{10}\ of\ the\ 250\ people\to\dfrac{3}{10}\cdot250=\dfrac{3}{1}\cdot25=75-number\ of\ boys\\\\100+75=175-number\ of\ girls\ and\ boys\\\\250-175=75-number\ of\ adults\\\\Answer:\ 75

3 0
3 years ago
What is the value of (-2.2) (2.2) (-1.6)
Firdavs [7]
-2.2*2.2*-1.6=7.744
Hope this helps!
3 0
3 years ago
Read 2 more answers
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