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Alenkasestr [34]
3 years ago
9

Y=1/2x what are the coordinates PLEASE HELP !!!!!!!

Mathematics
2 answers:
iris [78.8K]3 years ago
6 0
Y=1/2x represents a Linear equation from the form y = mx + b (In our case, b = 0 so we don't put it).
In such a form, m represents the slope of the equation. And b represents the y-intercept of the equation. While y and x represents the coordinates of 2 points.
In y=1/2x, m=1/2; b = 0. So the coordinates are y and x.
I've added a graph of the equation y= 1/2x under the answer.

Hope this Helps! :)

Gre4nikov [31]3 years ago
3 0
U need x also srry u cant complete ure answer

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y = 25x+100

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How many sides does a regular polygon have if each exterior angle has a measure of 15 degrees? (Show work!!)
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3 years ago
What inequality represents the sentence? The product of a number and 12 is no more than 15.
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7 0
3 years ago
im offering 100 points for this. Using the GCF you found in Part B, rewrite 56 + 96 as two factors. One factor is the GCF and th
Dennis_Churaev [7]

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7 0
2 years ago
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In 2010, the world's largest pumpkin weighed 1,810 kilograms. An average-sized pumpkin weighs 5,000 grams.
jekas [21]

Answer:

1,805 kg.

Step-by-step explanation:

We have been given that in 2010, the world's largest pumpkin weighed 1,810 kilograms. An average-sized pumpkin weighs 5,000 grams. We are asked to find the how much the world's largest pumpkin weighs than an average pumpkin.

First of all, we will convert the weight of average pumpkin in kilograms by dividing 5,000 by 1000 as 1 kg equals 1,000 gm.

\text{The weight of average pumpkin}=\frac{5000\text{ gm}}{\frac{\text{1000 gm}}{\text{ 1 kg}}}

\text{The weight of average pumpkin}=5000\text{ gm}\times \frac{\text{ 1 kg}}{\text{1000 gm}}

\text{The weight of average pumpkin}=5\text{ kg}

Now, we will subtract the weight of average pumpkin from world's largest pumpkin's weight.

1,810\text{ kilograms}-5\text{ kilograms}=1,805\text{ kilograms}

Therefore, the 2010 world-record pumpkin weighs 1,805 kilograms more than an average-sized pumpkin.

3 0
3 years ago
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