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3 years ago

List the range (1, -2), (-2,0), (-1,2), (1,3)

Mathematics

1 answer:

Vinil7 [7]3 years ago

5 0

Answer:

-2,0,2,3

Step-by-step explanation:

In this problem, the Domain are the first number in the parenthesis, while the range is the second.

1. (1, -2) - - > The second number is -2

2.(-2, 0) - - > The second number is 0

3. (-1,2) - - > The second number is 2

4. (1,3) - - > The second number is 3

Therefore the answers are -2,0,2,3

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Write a quadratic function f whose zeros are 13 and 2.

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Answer:

(x-13)(x+3) = f(x) = x^2 -10x -39

Step-by-step explanation:

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3 years ago

Use the given information to find (a) sin(s+t), (b) tan(s+t), and (c) the quadrant of s+t. cos s = - 12/13 and sin t = 4/5, s an

Anton [14]

Answer:

Part a) $sin(s + t) =-\frac{63}{65}$

Part b) $tan(s + t) = -\frac{63}{16}$

Part c) (s+t) lie on Quadrant IV

Step-by-step explanation:

[Part a) Find sin(s+t)

we know that

$sin(s + t) = sin(s) cos(t) + sin(t)cos(s)$

step 1

Find sin(s)

$sin^{2}(s)+cos^{2}(s)=1$

we have

$cos(s)=-\frac{12}{13}$

substitute

$sin^{2}(s)+(-\frac{12}{13})^{2}=1$

$sin^{2}(s)+(\frac{144}{169})=1$

$sin^{2}(s)=1-(\frac{144}{169})$

$sin^{2}(s)=(\frac{25}{169})$

$sin(s)=\frac{5}{13}$ ---> is positive because s lie on II Quadrant

step 2

Find cos(t)

$sin^{2}(t)+cos^{2}(t)=1$

we have

$sin(t)=\frac{4}{5}$

substitute

$(\frac{4}{5})^{2}+cos^{2}(t)=1$

$(\frac{16}{25})+cos^{2}(t)=1$

$cos^{2}(t)=1-(\frac{16}{25})$

$cos^{2}(t)=\frac{9}{25}$

$cos(t)=-\frac{3}{5}$ is negative because t lie on II Quadrant

step 3

Find sin(s+t)

$sin(s + t) = sin(s) cos(t) + sin(t)cos(s)$

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

substitute the values

$sin(s + t) = (\frac{5}{13})(-\frac{3}{5}) + (\frac{4}{5})(-\frac{12}{13})$

$sin(s + t) = -(\frac{15}{65}) -(\frac{48}{65})$

$sin(s + t) =-\frac{63}{65}$

Part b) Find tan(s+t)

we know that

tex]tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))[/tex]

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

step 1

Find tan(s)

$tan(s)=sin(s)/cos(s)$

substitute

$tan(s)=(\frac{5}{13})/(-\frac{12}{13})=-\frac{5}{12}$

step 2

Find tan(t)

$tan(t)=sin(t)/cos(t)$

substitute

$tan(t)=(\frac{4}{5})/(-\frac{3}{5})=-\frac{4}{3}$

step 3

Find tan(s+t)

$tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))$

substitute the values

$tan(s + t) = (-\frac{5}{12} -\frac{4}{3})/(1 - (-\frac{5}{12})(-\frac{4}{3}))$

$tan(s + t) = (-\frac{21}{12})/(1 - \frac{20}{36})$

$tan(s + t) = (-\frac{21}{12})/(\frac{16}{36})$

$tan(s + t) = -\frac{63}{16}$

Part c) Quadrant of s+t

we know that

$sin(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

$tan(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

Find the value of cos(s+t)

$cos(s+t) = cos(s) cos(t) -sin (s) sin(t)$

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

substitute

$cos(s+t) = (-\frac{12}{13})(-\frac{3}{5})-(\frac{5}{13})(\frac{4}{5})$

$cos(s+t) = (\frac{36}{65})-(\frac{20}{65})$

$cos(s+t) =\frac{16}{65}$

we have that

$cos(s+t)=positive$ -----> (s+t) could be in I or IV quadrant

$sin(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

$tan(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

therefore

(s+t) lie on Quadrant IV

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3 years ago

Please help with this question algebra 1

leva [86]

Answer:

Examples for the problem would be:

1) Robert started to run out of his house as fast as he can
2) But he slowed down when he met up with his friends
3) They then stopped at a park along the way
4) After stopping at the park, they ran the rest of the way to school hoping they wouldn't be late.
or
1) Robert met with his friends right outside his house and raced them to the next block
2) After Robert won, they all walked normally to the park
3) They then stopped at the park for a quick break
4) After realizing class was gonna start in a few minutes, they ran as fast as they could to school

Step-by-step explanation:

For the first part, you want Robert to be running at a normal running speed, such as him racing his friends or just running.
For the second part you want him to be walking at a slower pace.
For the third part you want them to not be moving
For the fourth part, you'll explain how/why they're moving so fast since the graph shows that he ran fastest at this time.

8 0

3 years ago

If ∠A and ∠B are vertical angles, and ∠A = 98°, find ∠B.

Arturiano [62]

Answer: [D]: 98°<span> .
______________________________________
Explanation:
______________________________________
Since vertical angles are congruent;
and since </span><span>∠A and ∠B are vertical angles;
then: m</span>∡A = m∡B .
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So given: m∡A = 98° ; then m∡B = 98° .
______________________________________

4 0

3 years ago