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3 years ago

A hubcap has a redius of 16 centimeter.What is the area of the hubcap?

1 answer:

6 0

Area=πr^2
Area=π(16)^2
Area≈804.2477 cm^2

I left the number large enough in case you need to round it off to some other decimal place.

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Which graph represents the solution set to this system of equations? –x + 2y = 6 and 4x + y = 3 On a coordinate plane, a line go

erastova [34]

Answer:

On a coordinate plane, a line goes through (0, 3) and (2, 4) and another line goes through (0, 3) and (0.75, 0).

This answer almost coincide with option C. I suppose there was a mistype.

Step-by-step explanation:

The system of equations is formed by:

–x + 2y = 6

4x + y = 3

In the picture attached, the solution set is shown.

The first equation goes through (0, 3) and (2, 4), as can be checked by:

–(0) + 2(3) = 6

–(2) + 2(4) = 6

The second goes through (0, 3) and (0.75, 0), as can be checked by:

4(0) + (3) = 3

4(0.75) + (0) = 3

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Between which two integers does √700 lie?

erastovalidia [21]

The square root of 700 is about 26.45, so it lies between 26 and 27

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Which ordered pair is a solution to the equation y=-2x+10

wolverine [178]

Are there any answer choices?

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<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{

Jlenok [28]

Answer:

$x>2$

Step-by-step explanation:

When given the following inequality;

$(x^2+x-3):(x^2-4)\geq1$

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

$\frac{x^2+x-3}{x^2-4}\geq1$

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

$\frac{x^2+x-3}{x^2-4}-1\geq0$

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

$\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0$

Perform the operation on the other side distribute the negative sign and combine like terms;

$\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0$

Factor the equation so that one can find the intervales where the inequality is true;

$\frac{x+1}{(x-2)(x+2)}\geq0$

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

$-1, 2, -2$

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

$x\leq-2\\-2$

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

$x\leq-2$ -> negative

$-2$ -> neagtive

$-1\leq x$ -> neagtive

$x>2$ -> positive

Therefore, the solution to the inequality is the following;

$x>2$

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3 years ago