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3 years ago

Solve and write in interval notation. x ≥ 2 and x < -3

1 answer:

8 0

Think: all values of x = to or > than 2 are ok, as are all values of x smaller than -3.

The first inequality, graphed, looks like "a solid dot at x = 2" and an arrow pointing to the right from that dot, plus an open circle at x = -3 and an arrow pointing to the left from that circle. These two arrows have NO overlap, so there is no "solution."

If you want the problem in interval notation, it's (-infinity, -3) ∪ [2, infinity).

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Whats is 67.24 as a mixed number

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Find the value of the logarithm Log 182 round answer to the nearest thousandth

Tresset [83]

$log(182)=log(2*7*13)$

As, $log(m*n)=log(m)+log(n)$

$log(2*7*13)=log(2)+log(7)+log(13)$

=0.301029+0.845098+1.13943

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Why do we use variables in math?

Phoenix [80]

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We use it to substitute an unknown number. We have a variable for a number we don’t know and as we solve the problem we can find the answer for the variable or unknown number.

Step-by-step explanation:

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1 year ago

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I really need help quick I’ve been working this out for 1 hour already

wolverine [178]

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Step-by-step explanation:

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2 years ago

A ball is projected in to the air. Its height at time t is given by the equation

Ksju [112]

The equation gives the height of the ball. That is, h is the height of the ball. t is the time. Since we are looking for the time at which the height is 8 (h=8), we need to set the equation equal to 8 and solve for t. We do this as follows:

$h=-16 t^{2} +60t+1$
$8=-16 t^{2} +60t+1$
$16 t^{2} -60t-1+8=0$
$16 t^{2} -60t+7=0$

This is a quadratic equation and as it is set equal to 0 we can solve it using the quadratic formula. That formula is:
$t= \frac{-bplus minus \sqrt{ b^{2}-4ac } }{2a}$
You might recall seeing this as "x=..." but since our equation is in terms of t we use "t-=..."

In order to use the formula we need to identify a, b and c.
a = the coefficient (number in front of) $t^{2}$ = 16.
b = the coefficient of t = -60
c = the constant (the number that is by itself) = 7

Substituting these into the quadratic formula gives us:
$t= \frac{-(-60)plus minus \sqrt{ (-60)^{2}-4(16)(7) } }{2(16)}$
$t= \frac{60plus minus \sqrt{3600-448 } }{32}$
$t= \frac{60plus minus \sqrt{3152 } }{32}$

As we have "plus minus" (this is usually written in symbols with a plus sign over a minus sign) we split the equation in two and obtain:
$t= \frac{60+56.1426}{32} =3.63$
and
$t= \frac{60-56.1426}{32} =.12$

So the height is 8 feet at t = 3.63 and t=.12

It should make sense that there are two times. The ball goes up, reaches it's highest height and then comes back down. As such the height will be 8 at some point on the way up and also at some point on the way down.

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3 years ago