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____ [38]
2 years ago
13

Expand the expression / brackets: 2y (3y + 9)=

Mathematics
1 answer:
Alecsey [184]2 years ago
7 0

Answer:

6y^2 + 18y

Step-by-step explanation:

2y (3y + 9)=

6y^2 + 18y

[2y (3y becomes 6y^2]

[2x(      + 9) becomes 18y]

Add them to obtain 6y^2 + 18y

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4 cot theta<br>=<br>tan theta<br>how do I I find the value of quadrant?​
Ray Of Light [21]

Answer:

4cotα=tanα

4(1/tanα)=tanα

(4/tanα)=tanα

cross multiply

=> 4=tan²α

√4=√tan²α

±2=tanα

α=arc( tan) |2|

α=63.4° ( in first quadrant)

and

α=180+63.4=243.4 in the third quadrant

since we also found a negative answer( i.e –2) then α also lies in quadrants where it gives a negative value(i.e second and fourth quadrants)

α=180–63.4=116.6° in the second quadrant

α=360–63.4=296.6 in the fourth quadrant

therefore theta( in my case, alpha) lies in all four quadrants and is equal to:

α=63.4°,243.4°,116.6°and 296.6°

5 0
4 years ago
15 POINTS PLEASE HELPPP :D
zhuklara [117]

Answer:

The equation is 1.96- 0.75

The answer is 1.21

Step-by-step explanation:

Hope this helps :)

4 0
3 years ago
Can someone help me
jeka57 [31]
OK SURE WHAT IS IT? SO I CAN ANSWER YOUR QUEST.?
5 0
3 years ago
We are interested in the amount that students study per week. Suppose you collected the following data in hours {4.4, 5.2, 6.4,
olga nikolaevna [1]

Answer:

Step-by-step explanation:

Hello!

The objective is to estimate the average time a student studies per week.

A sample of 8 students was taken and the time they spent studying in one week was recorded.

4.4, 5.2, 6.4, 6.8, 7.1, 7.3, 8.3, 8.4

n= 8

X[bar]= ∑X/n= 53.9/8= 6.7375 ≅ 6.74

S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/7*[376.75-(53.9²)/8]= 1.94

S= 1.39

Assuming that the variable "weekly time a student spends studying" has a normal distribution, since the sample is small, the statistic to use to perform the estimation is the student's t, the formula for the interval is:

X[bar] ± t_{n-1;1-\alpha /2}* (S/√n)

t_{n-1;1-\alpha /2}= t_{7;0.975}= 2.365

6.74 ± 2.365 * (1.36/√8)

[5.6;7.88]

Using a confidence level of 95% you'd expect that the average time a student spends studying per week is contained by the interval [5.6;7.88]

I hope this helps!

3 0
4 years ago
Read 2 more answers
The graph below represents the height of the ball with respect to time
IgorLugansk [536]
I don’t know I can’t help you
3 0
4 years ago
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