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just olya [345]

2 years ago

8

Which of the following sets of numbers could not represent the three sides of a right triangle? \{9,12,15\}{9,12,15} \{40,75,85\

}{40,75,85} \{61,63,87\}{61,63,87} \{14,48,50\}{14,48,50}

1 answer:

AlekseyPX2 years ago

3 0

Answer:

40,75,80

It is this because there is two sets of this number and that why

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What is log b^b^6x equivalent to (The b is under log and ^b)? How do you know?

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Work Shown:

For each step, the logs are all base b. This is to save time and hassle of writing tricky notation of having to write the smaller subscript 'b' multiple times. The first rule to use is that log(x^y) = y*log(x) for any base of a logarithm. The second rule is that $\log_b(b) = 1$ meaning that the log base of itself is 1

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2 years ago

the area of a gymnasium floor is 264 square yard. the floor is 11 yards wide . how long is the floor?

Alekssandra [29.7K]

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2 years ago

Read 2 more answers

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

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

But, angle $t$ is in 3rd quadrant, so value of

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

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2 years ago

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