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3 years ago

Helppppp.. picture attached

Mathematics

1 answer:

tatyana61 [14]3 years ago

6 0

$n^2+n\\\\check:\\\\a_1=1^2+1=2\\a_2=2^2+2=6\\a_3=3^2+3=12\\a_4=4^2+4=20$

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Can someone help me ?

lozanna [386]

Answer:

10 > v (if you don't trust me, check it yourself. Subsitute v for 10 in the first equation.

Step-by-step explanation:

1) 7 > v - 3

2) add 3 to both sides

3) so 7 plus 3 equals 10

So you basically do this

7 > v - 3 (ad 3 to both sides)

You get 10 > v (which is as simplified as possible)

Hope this helps! :)

8 0

3 years ago

write a ratio for the following description: for every 6 cups of flour in a bread recipe there are two cups of milk

8090 [49]

6:2 because there are 6 cups of flour and 2 cups of milk if u put the 2 first it means 2 cups to 6 cups and you need to answer it the exact way the it is asked I am in elementary school and I know this

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2 years ago

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I'm a little bit confused about this question.

Virty [35]

Answer:

Step-by-step explanation:

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3 years ago

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Out of a group of 120 students that were surveyed about winter sports 28 said they ski 53 said they snowboard Sixteen of the stu

hjlf

Answer:

P ( snowboard I ski) = 0.5714

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Ski

Event B: Snowboard.

28 out of 120 students ski:

This means that $P(A) = \frac{28}{120} = 0.2333$

16 out of 120 do both:

This means that $P(A \cap B) = \frac{16}{120} = 0.1333$

P ( snowboard I ski)

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1333}{0.2333} = 0.5714$

P ( snowboard I ski) = 0.5714

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3 years ago

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Suppose that the terminal side of angle alphaα lies in Quadrant I and the terminal side of angle betaβ lies in Quadrant IV. If s

melamori03 [73]

Solution :

It is given that :

$\alpha$ lies in the first quadrant.

And $\beta$ lies in the fourth quadrant.

Since, $\sin \alpha = \frac{5}{13}$ and $\cos \beta = \frac{6}{\sqrt{85}}$ (given)

$\sin \alpha = \frac{5}{13}$

$\cos \alpha = \sqrt{1-\sin^2 \alpha}$

$\cos \alpha = \frac{12}{13}$

Similarly $\cos \beta = \frac{6}{\sqrt{85}}$

$\sin \beta = \sqrt{1-\cos^2 \beta}$

$\sin \beta = \sqrt{1-\frac{36}{85}}$

$-\frac{7}{\sqrt{85}}$ (IVth quadrant)

Therefore,

$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$=\frac{12}{13}\times \frac{6}{\sqrt{85}}-\frac{5}{13}\times \frac{-7}{\sqrt{85}}$

$= \frac{107}{13 \sqrt{85}}$

6 0

2 years ago