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3 years ago

13

Olive Garden is only 10 square yards in the watermelon plants she wants to grow require two and a half square yards each how man

y watermelon plants can she grow

2 answers:

inysia [295]3 years ago

6 0

10 squared is 100 and 2.5 squared is 6.25. So if you divide 100 by 6.25 you get 16. So your answer is 16 watermelons.

aleksandr82 [10.1K]3 years ago

5 0

4 because 2.5 x 4 = 10 :)

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you are going to give each customer 12fl oz of lemonade.If you have 15 gallons of lemonade prepared, how many customers will you

zimovet [89]

Answer:

160

Step-by-step explanation:

so 1 gallon is 128 fluid ounces

how many is 12fl oz

so (1×12)/128=0.09375gallons per person

and you have 15 so if one consumes 0.09375 how many can consume 15

so 15/0.09375

8 0

3 years ago

35% of 20 is what number

EleoNora [17]

It's 7

Hope I helped! ( Smiles )

8 0

3 years ago

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The mean mass of five boys is 10.5 kg. The mean mass of three girls is 9.5 kg. Work out the mean mass of the eight young childre

11111nata11111 [884]

Answer:

the mean of 8 children = mean of 5 boys + mean of 3 girls / 2

10.5 + 9.5 / 2

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= 10 kg

therefore mean of 8 children is 10 kg

3 0

3 years ago

Fine the slope of the line that passes through (9,3) (2,2)

Charra [1.4K]

Just substitute it into the gradient formula which is: y2 - y1 / x2 - x1. So 2-3/2-9 = -1/-7 = 1/7

4 0

3 years ago

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Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea

Gennadij [26K]

Answer:

$y_g(t) = c_1*( 2t - 1 ) + c_2*e^(^-^2^t^) - e^(^-^2^t^)* [ t^3 + \frac{3}{4}t^2 + \frac{3}{4}t ]$

Step-by-step explanation:

Solution:-

- Given is the 2nd order linear ODE as follows:

$ty'' + ( 2t - 1 )*y' - 2y = 6t^2 . e^(^-^2^t^)$

- The complementary two independent solution to the homogeneous 2nd order linear ODE are given as follows:

$y_1(t) = 2t - 1\\\\y_2 (t ) = e^-^2^t$

- The particular solution ( yp ) to the non-homogeneous 2nd order linear ODE is expressed as:

$y_p(t) = u_1(t)*y_1(t) + u_2(t)*y_2(t)$

Where,

$u_1(t) , u_2(t)$ are linearly independent functions of parameter ( t )

- To determine [ $u_1(t) , u_2(t)$ ], we will employ the use of wronskian ( W ).

- The functions [ $u_1(t) , u_2(t)$ ] are defined as:

$u_1(t) = - \int {\frac{F(t). y_2(t)}{W [ y_1(t) , y_2(t) ]} } \, dt \\\\u_2(t) = \int {\frac{F(t). y_1(t)}{W [ y_1(t) , y_2(t) ]} } \, dt \\$

Where,

F(t): Non-homogeneous part of the ODE

W [ y1(t) , y2(t) ]: the wronskian of independent complementary solutions

- To compute the wronskian W [ y1(t) , y2(t) ] we will follow the procedure to find the determinant of the matrix below:

$W [ y_1 ( t ) , y_2(t) ] = | \left[\begin{array}{cc}y_1(t)&y_2(t)\\y'_1(t)&y'_2(t)\end{array}\right] |$

$W [ (2t-1) , (e^-^2^t) ] = | \left[\begin{array}{cc}2t - 1&e^-^2^t\\2&-2e^-^2^t\end{array}\right] |\\\\W [ (2t-1) , (e^-^2^t) ]= [ (2t - 1 ) * (-2e^-^2^t) - ( e^-^2^t ) * (2 ) ]\\\\W [ (2t-1) , (e^-^2^t) ] = [ -4t*e^-^2^t ]\\$

- Now we will evaluate function. Using the relation given for u1(t) we have:

$u_1 (t ) = - \int {\frac{6t^2*e^(^-^2^t^) . ( e^-^2^t)}{-4t*e^(^-^2^t^)} } \, dt\\\\u_1 (t ) = \frac{3}{2} \int [ t*e^(^-^2^t^) ] \, dt\\\\u_1 (t ) = \frac{3}{2}* [ ( -\frac{1}{2} t*e^(^-^2^t^) - \int {( -\frac{1}{2}*e^(^-^2^t^) )} \, dt] \\\\u_1 (t ) = -e^(^-^2^t^)* [ ( \frac{3}{4} t + \frac{3}{8} )] \\\\$

- Similarly for the function u2(t):

$u_2 (t ) = \int {\frac{6t^2*e^(^-^2^t^) . ( 2t-1)}{-4t*e^(^-^2^t^)} } \, dt\\\\u_2 (t ) = -\frac{3}{2} \int [2t^2 -t ] \, dt\\\\u_2 (t ) = -\frac{3}{2}* [\frac{2}{3}t^3 - \frac{1}{2}t^2 ] \\\\u_2 (t ) = t^2 [\frac{3}{4} - t ]$

- We can now express the particular solution ( yp ) in the form expressed initially:

$y_p(t) = -e^(^-^2^t^)* [\frac{3}{2}t^2 + \frac{3}{4}t - \frac{3}{8} ] + e^(^-^2^t^)*[\frac{3}{4}t^2 - t^3 ]\\\\y_p(t) = -e^(^-^2^t^)* [t^3 + \frac{3}{4}t^2 + \frac{3}{4}t - \frac{3}{8} ] \\$

Where the term: 3/8 e^(-2t) is common to both complementary and particular solution; hence, dependent term is excluded from general solution.

- The general solution is the superposition of complementary and particular solution as follows:

$y_g(t) = y_c(t) + y_p(t)\\\\y_g(t) = c_1*( 2t - 1 ) + c_2*e^(^-^2^t^) - e^(^-^2^t^)* [ t^3 + \frac{3}{4}t^2 + \frac{3}{4}t ]$

3 0

3 years ago