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2 years ago

The perimeter of the rectangle below is 222 units find the length of side RS

Mathematics

2 answers:

timurjin [86]2 years ago

7 0

Answer:

111

Step-by-step explanation:

divide 222 by 2 and you get your answer

34kurt2 years ago

4 0

Answer:

55.5

Step-by-step explanation:

since we don't know the length or width of any side, one length side will only be 1/4 of 222. And perimeter is all the sides added up. This rectangle will actually be a square.

222/4=55.5

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Lydia is a childbirth assistant. The first year of her career, she attended the births of 2 babies. Every year after that, the n

Sonbull [250]

Answer:24 or 4

Step-by-step explanation:

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3 years ago

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kvv77 [185]

It would be a 1 in 4 chance

8 0

3 years ago

Please help ASAP!

Kruka [31]

First, recall that

$\sin(-\theta)=-\sin\theta$

So if $\sin(-\theta)=\dfrac15$ , then $\sin\theta=-\dfrac15$ .

Second,

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

We know that $\tan\theta>0$ and $\sin\theta$ , which means we should also have $\cos\theta$ .

Third,

$\sin^2\theta+\cos^2\theta=1\implies\cos\theta=\pm\sqrt{1-\sin^2\theta}$

but as we've already shown, we need to have $\cos\theta$ , so we pick the negative root.

Finally,

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}\iff\dfrac6{\sqrt{12}}=\dfrac{-\frac15}{\cos\theta}\implies\cos\theta=-\dfrac1{5\sqrt3}$

Unfortunately, none of the given answers match, so perhaps I've misunderstood one of the given conditions... In any case, this answer should tell you everything you need to know to find the right solution from the given options.

6 0

3 years ago

Solve the following recurrence relation: <br> <img src="https://tex.z-dn.net/?f=A_%7Bn%7D%3Da_%7Bn-1%7D%2Bn%3B%20a_%7B1%7D%20%3D

-Dominant- [34]

By iteratively substituting, we have

$a_n = a_{n-1} + n$

$a_{n-1} = a_{n-2} + (n - 1) \implies a_n = a_{n-2} + n + (n - 1)$

$a_{n-2} = a_{n-3} + (n - 2) \implies a_n = a_{n-3} + n + (n - 1) + (n - 2)$

and the pattern continues down to the first term $a_1=0$ ,

$a_n = a_{n - (n - 1)} + n + (n - 1) + (n - 2) + \cdots + (n - (n - 2))$

$\implies a_n = a_1 + \displaystyle \sum_{k=0}^{n-2} (n - k)$

$\implies a_n = \displaystyle n \sum_{k=0}^{n-2} 1 - \sum_{k=0}^{n-2} k$

Recall the formulas

$\displaystyle \sum_{n=1}^N 1 = N$

$\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2$

It follows that

$a_n = n (n - 2) - \dfrac{(n-2)(n-1)}2$

$\implies a_n = \dfrac12 n^2 + \dfrac12 n - 1$

$\implies \boxed{a_n = \dfrac{(n+2)(n-1)}2}$

4 0

2 years ago

The point-slope form of the equation of a line that passes through points (8, 4) and (0, 2) is y – 4 = y minus 4 equals StartFra

Stels [109]

Answer:

Third option: $y= \frac{1}{4}x+2$

Step-by-step explanation:

<h3><em> The correct form of the exercise is: "The point-slope form of the equation of a line that passes through points (8, 4) and (0, 2) is $y -4 = \frac{1}{4}(x -8)$ . What is the slope-intercept form of the equation for this line?"</em></h3><h3><em /></h3>

<em> </em>The equation of the line in Slope-Intercept form is:

$y=mx+b$