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2 years ago

Suppose five pumpkins are weighed two at a time in all possible ways. The weights in pounds are recorded as 16,18,19,20,21,22,23

,24,26, and 27. How much does each individual pumpkin weigh?

Mathematics

1 answer:

monitta2 years ago

5 0

$a+b=16 => \boxed{a=16-b} \\ \\ b+c=18=>\boxed{c=18-b} \\ \\c+d=19 \\ \\ a+c=21 \\ \\ d+e=20 \\ \\ a+d=22 \\ \\ a+e=23 \\ \\ b+d=24 \\ \\ b+e=26 \\ \\ \\ a+c=21 \\ 16-b+18-b=21 \\ 32-2b=21 \\ -2b=21-34 \\ -2b=-13 \\ \\ \boxed{b=\frac{-13}{-2}=6.5} \\ \\ a=16-6.5 \\ \\ \boxed{a=9.5} \\ \\ c=18-6.5 \\ \\ \boxed{c=11.5} \\ \\ d=19-11.5 \\ \\ \boxed{d=7.5} \\ \\ e=20-7.5 \\ \\ \boxed{e=12.5}$

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Justin bought some yarn from his favorite craft store. He can make one scarf with 3/5 of a ball of yarn. If he purchases 15 ball

Elena L [17]

He can make 25 scarves from 15 balls of yarn.

Step-by-step explanation:

Given,

Yarn used for one scarf = $\frac{3}{5}$ of a ball

Yarn purchased = 15 balls

Let,

x be the number of scarves.

Yarn used for one scarf*number of scarves = Yarn purchased

$\frac{3}{5}x=15$

Multiplying both sides by $\frac{5}{3}$

$\frac{5}{3}*\frac{3}{5}x=15*\frac{5}{3}\\x=\frac{75}{3}\\x=25$

He can make 25 scarves from 15 balls of yarn.

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2 years ago

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2 years ago

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faust18 [17]

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2 years ago

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Andrew bought 7 bundles of drawing sheets that each cost the same amount of money. The tax on his purchase was $1. He spent a to

Annette [7]

Solve the equation 7x-1=15 where x is the cost of each bundle of drawing sheets.

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2 years ago

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nignag [31]

Answer:

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

Step-by-step explanation:

$\int\frac{x^{4}}{x^{4} -1}dx$

Adding and Subtracting 1 to the Numerator

$\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx$

Dividing Numerator seperately by $x^{4} - 1$

$\int 1 + \frac{1}{x^{4}-1 }\, dx$

Here integral of 1 is x +c1 (where c1 is constant of integration

$x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx$ ----------------------------------(1)

We apply method of partial fractions to perform the integral

$\frac{1}{(x-1)(x+1)(x^{2}+1)}$ = $\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}$ ------------------------------------------(2)

$\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}$

1 = $A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)$ -------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = $\frac{1}{4}$

1 = B(-1-1)(1+1)

B = $-\frac{1}{4}$

1 = C(i-1)(i+1)

C = $-\frac{1}{2}$

Substituting A, B, C in equation (2)

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }$

On integration

Here $\int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx$

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\frac{1}{4} ln(x-1)$ - $\frac{1}{4} ln(x+1)$ - $\frac{1}{2} arctanx$ + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

$x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2$

Here c1 + c2 can be added to another and written as c

Therefore,

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

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2 years ago