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Zolol [24]
3 years ago
6

Given, Solve the trigonometric equation for . Express in radians only.

Mathematics
1 answer:
natulia [17]3 years ago
5 0
Umm, what’s the equation?
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seraphim [82]
1. 2x and 6y Because
y = 2x + 6
4 0
2 years ago
Write the correct answer:
natta225 [31]

Answer:

-1 for question 1, -11 for question 2, 7 for question 3

Step-by-step explanation:

8 0
3 years ago
What conversions need to be performed to find the number of drops in 20ml
siniylev [52]
1\ drop\approx65\mu l\\\\1ml=1000\mu l\\20ml=20\cdot1000\mu l=20,000\mu l\\\\20,000\mu l:65\mu l\approx308\ (drops)
5 0
3 years ago
The ratio of the areas of corresponding faces of two cubes is 1:16. what is the ratio of their volumes?
Yakvenalex [24]
The ratio of the areas is the one-to-one measure squared.  In order to find the ratio of their volumes, which is a cubed measure, we have to find the one-to-one and them cube it.  The square root of 1/16 is 1/4.  Now if we cube 1/4 to get the volume, it will be 1/64. 
6 0
3 years ago
How do you do this question?
alina1380 [7]

Answer:

(8√2) / 15

Step-by-step explanation:

A curve bounded by the y-axis is represented by in terms of dy;

\int \:x\:dt

When the curve crosses the y-axis, x will be 0. In this case x is the function of t, so we have to solve for x(t) = 0;

0 = t^2 + 2t --- (1)

Solution(s) => t = 0, t = 2

dy = (1/2 * 1/√t)dt --- (2)

Our solutions (0, 2) are our limits. The area of the curve is in the form A\:=\:\int _b^a\:f\left(t\right)g'\left(t\right)dt , so now let's introduce the limits of integration, x(t) and dy/dt. Remember, dy/dt = (1/2 * 1/√t) (second equation). 1/2 * 1/√t can be rewritten as 1/2 * t^(-1/2)....

A\:=\:\int _2^0\:\left(t^2-2t\right)\left(\frac{1}{2}t^{-\frac{1}{2}}\right)dt\\\\= \int _2^0\:\left(\frac{1}{2}t^{\frac{3}{2}}-t^{\frac{1}{2}}\right)dt\\\\= \left[\frac{t^{\frac{5}{2}}}{5}-\frac{2t^{\frac{3}{2}}}{3}\right]_2^0\\\\= 0\:-\:\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\\\\= \frac{8\sqrt{2}}{15}

Your solution is 8√2 / 15

7 0
2 years ago
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