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2 years ago

After I substitute in y how do I determine the coordinate(s)?

Mathematics

1 answer:

klasskru [66]2 years ago

6 0

Simplifying both sides of the equation then isolating the variable

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I really need help with this

postnew [5]

Answer:

1/2 or .5

Step-by-step explanation:

you find 2 points that the line goes through, then you find the rise/run. lastly you simplify if you can, then you have the answer

7 0

2 years ago

This table shows the ratio of number of people to liters of ginger ale needed find the missing values

makvit [3.9K]

Answer:

2 and 6

Step-by-step explanation:

7 0

2 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

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

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

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

Then

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

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

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

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

1 year ago

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

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

4 0

2 years ago

Please help I have been stuck for a while

goldfiish [28.3K]

Answer:

B. 50 degrees

Step-by-step explanation:

I am not 100% sure and if this is a right triangle, then it's measure is 90 degrees. So, 140 degrees minus 90 degrees would total, 50 degrees.

6 0

2 years ago

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