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dimaraw [331]
2 years ago
14

Jackson is three times as heavy as Andrew. If the sum of their weight is 360 pounds, find Andrews weight​

Mathematics
2 answers:
Elena-2011 [213]2 years ago
8 0

Answer:

40 pounds is the correct answer....

Hope this helps.......

Delicious77 [7]2 years ago
5 0
Answer: Andrew = 90 pounds

Explanation:
Assume Andrew weight is X
Since Jackson weight is 3 times Andrew
Then it’s 3x

3x + x = 360
4x = 360
x = 90
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Four friends are playing a game. They randomly choose a handful of marbles from a bag. The player with the highest ratio of blue
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We are given the number of Blue and Purple Marbles for each of the four friends. So first we have to find the ratio of blue marbles to purple marbles for each of them.

A) Rosa

Number of blue marbles = 10

Number of purple marbles = 12

Ratio of blue to purple marbles = \frac{10}{12}=0.833

B) Chi

Number of blue marbles = 5

Number of purple marbles = 6

Ratio of blue to purple marbles = \frac{5}{6}=0.833

C) Alberto

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Ratio of blue to purple marbles = \frac{5}{12}=0.417

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Ratio of blue to purple marbles = \frac{10}{6}=1.667

From the above calculations we can see that Pedro has the highest value of ratio for blue marbles to purple marbles. Also we can see Pedro is the only one for whom the number of blue marbles is more than the number of purple marbles. So this confirms that our calculation is correct.

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2 years ago
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The spring has a stiffness k=200n/m and is unstretched when the 25 kg block is at
g100num [7]
To solve this we are going to use the formula fro the force applied to a spring: F=ks
where
k is the spring constant 
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Since we know the F=ma, we can replace that in our formula and solve for a :
ma=ks
a= \frac{ks}{m}
where
a is the acceleration 
k is the spring constant
s is the extension 
m is the mass

We know for our problem that k=200, s=0.4, and m=25. So lets replace those values in our formula to find a:
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Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y
zalisa [80]

Answer:

y has a finite solution for any value y_0 ≠ 0.

Step-by-step explanation:

Given the differential equation

y' + y³ = 0

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dy/dx + y³ = 0

Multiplying through by dx

dy + y³dx = 0

Divide through by y³, we have

dy/y³ + dx = 0

dy/y³ = -dx

Integrating both sides

-1/(2y²) = - x + c

Multiplying through by -1, we have

1/(2y²) = x + C (Where C = -c)

Applying the initial condition y(0) = y_0, put x = 0, and y = y_0

1/(2y_0²) = 0 + C

C = 1/(2y_0²)

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2y² = 1/[x + 1/(2y_0²)]

y² = 1/[2x + 1/(y_0²)]

y = 1/[2x + 1/(y_0²)]½

This is the required solution to the initial value problem.

The interval of the solution depends on the value of y_0. There are infinitely many solutions for y_0 assumes a real number.

For y_0 = 0, the solution has an expression 1/0, which makes the solution infinite.

With this, y has a finite solution for any value y_0 ≠ 0.

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