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hram777 [196]
3 years ago
10

Which geometric solid is the best model for the head of a human being?

Mathematics
1 answer:
Tasya [4]3 years ago
5 0
Sphere, ur head is the closest shape to a sphere
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Geometry help needed.
Margaret [11]

A'B'C' is larger than ABC, so the scale factor would be a whole number, scale factor of 2.

A'B'C'is also a morrir image of ABC, notice on ABC A is on the left and B is on the right, but on A'B'C', B" is on the left and A' is on the right.

This means it was reflected across the Y axis.

The first answer is the correct one.

4 0
3 years ago
The equation must be in slope-intercept form ____________.
Genrish500 [490]
Slope intercept form is y=mx+b
you can later plug in values to make your own slope intercept form.

4 0
4 years ago
an ice machine makes 1 1/2 pounds of ice every 1/2 hour. witch graph shows this proportional relationship?​
Roman55 [17]

Answer:

The 4th one to the Right

Step-by-step explanation:

4 0
3 years ago
Pls give answer I need to submit today​
BabaBlast [244]

Answer:

a \: 5 \: 10 \: 15 \: 20 \: three \: of \: these \: numbers \\ b \: 2  < 4 < 6 < 8 < 10 < 12 < 14 < 16 <  \\ 18 < 20 \: three \: of \: these \: numbers \\ sum \: of \: numbers < 40 \\  \\ numbers \\ 10 \\ 20 \\ 5 \\ 2

7 0
3 years ago
Please help! I've already answered part a, I don't understand what part b is asking.
Luba_88 [7]

Step-by-step explanation:

So, there is something known as a removable discontinuity, and it's essentially where you can define f(x) using the most simplified fraction, where you could normally not define f(x).

So we have the following equation:

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

As you may know, we cannot divide a number by the value of zero. When the denominator is equal to zero, on the graph this will appear as a vertical asymptote, where x approaches the value that makes the denominator zero, but never actually reaches it.

If you look at each denominator, you can set them equal to zero to find the vertical asymptotes

x+1 = 0

x=-1

There should be a vertical asymptote at x=-1, since it would make two of the denominators equal to -1, but let's divide the two fractions first.

Original Equation

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

Keep, change, flip

f(x) = (\frac{x+5}{x+1}*\frac{(x-4)(x+1)}{(x+3)(x-2)})-\frac{1}{x-2}

Multiply the two fractions

f(x) = (\frac{(x+5)(x-4)(x+1)}{(x+1)(x+3)(x-2)})-\frac{1}{x-2}

Notice how the x+1 is in the numerator and fraction? That means we can cancel it out!

f(x) = (\frac{(x+5)(x-4)}{(x+3)(x-2)})-\frac{1}{x-2}

In this simplified version of the fraction, we can technically define f(-1), but in the original version, since it's not defined there is a removable discontinuity at x=-1, meaning there is no vertical asymptote, but the function is still not defined at f(-1), and there will be a hole at that point.

4 0
2 years ago
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