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slega [8]
3 years ago
12

Mrs. Khan is mailing 2 packages. The large package weighs 6 2 3 pounds. The small package weighs 3 1 2 pounds. Which statement B

EST describes how to find out how much heavier the large package is? A) Add 6 2 3 + 3 1 2 . B) Divide 6 2 3 ÷ 3 1 2 C) Subtract 6 2 3 – 3 1 2 . D) Multiply 6 2 3 × 3 1 2 .
Mathematics
2 answers:
vagabundo [1.1K]3 years ago
6 0

<u>Answer:</u>

Mrs. Khan is mailing 2 packages. Subtract 623 – 312 best describe how to find out how much heavier the larger package is. Hence Option C is correct

<u>Solution:</u>

Given that  

Mrs khan is mailing two packages.

Weight of larger package = 623 pounds  

Weight of small package = 312 pounds

Now we need to find how much heavier larger package compare to smaller one is.

To get this if we remove weight of smaller package from larger package , we will get how much larger package is heavier.

That is<em> Weight of larger package – Weight of smaller package</em> = 623 pounds – 312 pounds

Hence option C) subtract 623 – 312 best describe how to find out how much heavier the larger package is.

amid [387]3 years ago
3 0

Answer:

The answer is to subtract 6 2/3 - 3 1/2

Step-by-step explanation:

In order to find out how much heavier the larger package is, subtract 6

2

3

– 3

1

2

.

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Answer:

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Step-by-step explanation:

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6 0
4 years ago
The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9
mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

<em>The height </em>h(t)<em> of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation </em>h(t) = 3 +90\cdot t -16\cdot t^{2}<em>, where </em>t<em> is time in seconds. After how many seconds will the ball be 84 feet above the ground. </em>

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By Quadratic Formula:

t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}

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The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

3 0
3 years ago
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