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4 years ago

Mixed numbers and improper fractions are often the result of measurements or calculations.

Mathematics

2 answers:

Nitella [24]4 years ago

7 0

Answer:

I'm pretty sure it's false

Step-by-step explanation:

lesya [120]4 years ago

3 0

The answer will mostly be true if you know the answers

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The graphs of f(x) and g(x) are shown below:

Alekssandra [29.7K]

Your Answer: G(x) = (x − 5)²

I just finished my semester exam, and this was one of the questions. I guessed on this question because I had no clue of what the answer was, and lucky me, I got the question right :D

Hope this helps y'all with your test :)

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3 years ago

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The ages are 3, 3, and 8

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3 years ago

What is the first quartile, Q1, of the data represented by the box plot?

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In this box plot 12.5 would be your answer

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3 years ago

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A cube has side length a. The side lengths are decreased to 20% of their original size. Write an expression in simplest form for

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3 years ago

The endpoints of a diameter of a circle are (6,2) and (−2,5) . What is the standard form of the equation of this circle?

Paul [167]

Answer:

The standard form of the given circle is

$(x-2)^2+(y-\frac{7}{2})^2=73$

Step-by-step explanation:

Given that the end points of a diameter of a circle are (6,2) and (-2,5);

Now to find the standard form of the equation of this circle:

The center is (h,k) of the circle is the midpoint of the given diameter

midpoint formula is $M=(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

Let $(x_{1},y_{1})$ and $(x_{2},y_{2})$ be the given points (6,2) and (-2,5) respectively.

$M=(\frac{6-2}{2},\frac{2+5}{2})$

$M=(\frac{4}{2},\frac{7}{2})$

$M=(2,\frac{7}{2})$

Therefore the center (h,k) is $(2,\frac{7}{2})$

now to find the radius:

The diameter is the distance between the given points (6,2) and (-2,5)

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

$=\sqrt{(-2-6)^2+(5-2)^2}$

$=\sqrt{(-8)^2+(3)^2}$

$=\sqrt{64+9}$

$=\sqrt{73}$

Therefore the radius is $\sqrt{73}$

i.e., $r=\sqrt{73}$

Therefore the standard form of the circle is

$(x-h)^2+(y-k)^2=r^2$

Now substituting the center and radiuswe get

$(x-2)^2+(y-\frac{7}{2})^2=(\sqrt{73})^2$

$(x-2)^2+(y-\frac{7}{2})^2=73$

Therefore the standard form of the given circle is

$(x-2)^2+(y-\frac{7}{2})^2=73$

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3 years ago