7. The graph shows the height, h, in feet, of water

Mathematics

1 answer:

MariettaO [177]1 year ago

4 0

The domain is talking about the "range" of the horizontal axis therefore you will be focusing on the x-intercepts.

The answer will be All non-negative real numbers less than or equal to 18

because the x-intercepts lies at 0 and 18. The answer makes sense because the furthest you can go is 18 ft and the closest you could go is 0 ft. The "all non-negative real numbers" puts a restriction on the least distance it could travel so that means that it stops at 0 ft because if you go any further, you will end up in the negatives and it clearly states "non-negative".

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Given that Cosecant (t) = negative StartFraction 13 Over 5 EndFraction for Pi less-than t less-than StartFraction 3 pi Over 2 En

Sergio [31]

Answer:

$\bold{cot(t) =\dfrac{12}{5}}$

Step-by-step explanation:

Given that:

$Cosec (t) = -\frac{13}5$

for $\pi < t < \frac{3 \pi}2$

That means, angle $t$ is in the 3rd quadrant.

To find:

Value of cot(t)

Solution:

First of all, let us recall what trigonometric ratios are positive and what trigonometric ratios are negative in 3rd quadrant.

In 3rd quadrant, tangent and cotangent are positive.

All other trigonometric ratios are negative.

Let us have a look at the following identity:

$cosec^2\theta -cot^2\theta =1$

here, $\theta =t$

So, $cosec^2t-cot^2t=1$

$\Rightarrow (-\dfrac{13}{5})^2-cot^2t=1\\\Rightarrow (\dfrac{169}{25})-cot^2t=1\\\Rightarrow \dfrac{169}{25}-1=cot^2t\\\Rightarrow \dfrac{169-25}{25}=cot^2t\\\Rightarrow \dfrac{144}{25}=cot^2t\\\Rightarrow cot(t)=\pm\sqrt{\dfrac{144}{25}}\\\Rightarrow cot(t)=\pm\dfrac{12}{5}$