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2 years ago

If x = a sin teta and y = b cos teta

Mathematics

1 answer:

Marat540 [252]2 years ago

8 0

x= a sin theta
y=b sin theta

$\\ \sf\longmapsto b^2x^2+a^2y^2$

$\\ \sf\longmapsto b^2(asin\Theta)^2+a^2(bcos\Theta)^2$

$\\ \sf\longmapsto b^2a^2sin^2\Theta+a^2b^2cos^2\Theta$

$\\ \sf\longmapsto a^2b^2(sin^2\Theta+cos^2Theta)$

$\\ \sf\longmapsto a^2+b^2(1)$

$\\ \sf\longmapsto a^2b^2$

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Can someone help with this problem

WARRIOR [948]

35+25=60
y=60

If you add the opposite side values of the triangle (not the one touching the y), you get the value of y). 35+25=y
This works because y+x= 180 and the three angles of a triangle add to 180.

4 0

2 years ago

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Since 2010, when 102390 Cases were reported, each year the number of new flu cases decrease to 85% of the prior year. Predict th

BaLLatris [955]

Answer:

20,158 cases

Step-by-step explanation:

Let $t=0$ represent year 2010.

We have been given that since 2010, when 102390 Cases were reported, each year the number of new flu cases decrease to 85% of the prior year.

Since the flu cases decrease to 85% of the prior year, so the flu cases for every next year will be 85% of last year and decay rate is 15%.

We can represent this information in an exponential decay function as:

$F(t)=102,390(1-0.15)^t$

$F(t)=102,390(0.85)^t$

To find number of cases in 2020, we will substitute $t=10$ in our decay function as:

$F(10)=102,390(0.85)^{10}$

$F(10)=102,390(0.1968744043407227)$

$F(10)=20,157.970260446597\approx 20,158$

Therefore, 20,158 cases will be reported in 2020.

3 0

2 years ago

6 / 32 Simplify a) 4p + 2p – 3p b) 9s- 5s + 2s-S c) 8t + 3r - 76 - 9r

Charra [1.4K]

Answer:

a.3p

b 5s

c 2r -76 this can be the answer

3 0

2 years ago

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One sample has a mean of and a second sample has a mean of . The two samples are combined into a single set of scores. What is t

Murrr4er [49]

Answer:

a) For this case we can use the definition of weighted average given by:

$M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}$

And if we replace the values given we have:

$M = \frac{8*4 + 16*4}{4+4}= 12$

b) $M = \frac{8*3 + 16*5}{3+5}= 13$

c) $M = \frac{8*5 + 16*3}{5+3}= 11$

Step-by-step explanation:

Assuming the following question: "One sample has a mean of M=8 and a second sample has a mean of M=16 . The two samples are combined into a single set of scores.

a) What is the mean for the combined set if both of the original samples have n=4 scores "

For this case we can use the definition of weighted average given by:

$M = \frac{ \bar X_1 n_1 + \bar X_2 n_2}{n_1 +n_2}$

And if we replace the values given we have:

$M = \frac{8*4 + 16*4}{4+4}= 12$

b) what is the mean for the combined set if the first sample has n=3 and the second sample has n=5

Using the definition we have:

$M = \frac{8*3 + 16*5}{3+5}= 13$

c) what is the mean for the combined set if the first sample has n=5 and the second sample has n=3

Using the definition we have:

$M = \frac{8*5 + 16*3}{5+3}= 11$

7 0

2 years ago

3. approximate value of log 3 base 7

xxTIMURxx [149]

Answer:

0.56457

Step-by-step explanation:

log 3 with base 7 = (Log 3)/ (Log 7) = 0.47712125 / 0.845098040014

(Log 3) / (Log 7) = 0.56457

3 0

2 years ago