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marin [14]
3 years ago
10

The average weight of 3 containers A,B and C is 3.2kg. Container A is twice as heavy as container B. Containers B is 400 g heavi

er than container C. Find the weight of container C.
Mathematics
1 answer:
dlinn [17]3 years ago
4 0

Answer:

The weight of container C is 2.1kg.

Step-by-step explanation:

This question is solved using a system of equations.

I am going to say that:

x is the weight of container A.

y is the weight of container B.

z is the weight of container C.

The average weight of 3 containers A,B and C is 3.2kg.

This means that the total weight is 3*3.2 = 9.6kg. So

x + y + z = 9.6

Container A is twice as heavy as container B.

This means that x = 2y.

Containers B is 400 g heavier than container C.

400g is 0.4kg. So

This means that y = z + 0.4, or z = y - 0.4

Replacing y and z as functions of x in the first equation:

x + y + z = 9.6

2y + y + y - 0.4 = 9.6

4y = 10

y = \frac{10}{4} = 2.5

Container C

z = y - 0.4 = 2.5 - 0.4 = 2.1

The weight of container C is 2.1kg.

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Step-by-step explanation:

using formula

a^2+b^2=c^2

x^2+4^2=8^2

x^2+16=64

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3 0
3 years ago
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Jai bought 2 pounds of grapes for $2.49 per pound and 3 muffins for $ .75 per muffin. How much did he pay for the grapes and muf
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$7.23

Step-by-step explanation:

(2 x 2.49) + (3 x .75) = 4.98 + 2.25 = $7.23

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3 years ago
What is the exact distance from (–1, 4) to (6, –2)? (4 points) units units units units
irina1246 [14]

Answer:

\sqrt{85}

Step-by-step explanation:

Calculate the distance d using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 1, 4) and (x₂, y₂ ) = (6, - 2)

d = \sqrt{(6+1)^2+(-2-4)^2}

  = \sqrt{7^2+(-6)^2}

  = \sqrt{49+36}

  = \sqrt{85} ← exact distance

4 0
4 years ago
A fish was swimming LaTeX: 2\frac{1}{2}2 1 2 feet below the water's surface at 1:00 p.M. Three hours later, the fish was at a de
marin [14]

Answer:

\frac{23}{4} feet

Step-by-step explanation:

The depth of the fish to the surface of water at 1:00 pm = 2 \frac{1}{2} feet

                                                                             =   \frac{5}{2} feet

The depth of the fish to its initial position at 4:00 pm = 3 \frac{1}{4} feet

                                                                              = \frac{13}{4} feet

The position of the fish with respect to the water's surface at 4:00 pm = the depth of the fish to the surface of water at 1:00 pm + the depth of the fish to its initial position at 4:00 pm

                                        =  \frac{5}{2} + \frac{13}{4}

                                        = \frac{10 + 13}{4}

                                         = \frac{23}{4}

The position of the fish with respect to the water's surface at 4:00 pm is \frac{23}{4} feet.

8 0
3 years ago
Please answer quick!!
rewona [7]

Answer: B

Step-by-step explanation:

\frac{2}{7k}(k-7)

\frac{2k-14}{7k}

\frac{2k}{7k}-\frac{14}{7k}

\frac{2}{7}-\frac{2}{k}

7 0
3 years ago
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