Answer:
<h2>
The right option is twelve-fifths</h2>
Step-by-step explanation:
Given a right angle triangle ABC as shown in the diagram. If ∠BCA = 90°, the hypotenuse AB = 26, AC = 10 and BC = 24.
Using the SOH, CAH, TOA trigonometry identity, SInce we are to find tanA, we will use TOA. According to TOA;
Tan (A) = opp/adj
Taken BC as opposite side since it is facing angle A directly and AC as the adjacent;
tan(A) = BC/AC
tan(A) = 24/10
tan(A) = 12/5
The right option is therefore twelve-fifths
I would subtract the 2nd from the first one
so -4x - -2x = -2x
4y - 4y = 0
-20 - -2 = -18
-2x = -18
now I would divide that by -2 to get the value for x
so no we have x = 9
now you just have to substitute that into one of the equations and simplify to get y
-2 X 9 + 4y = -2
-18 + 4y = -2
4y =16
y = 4
The pool is 40 since half of the other rectangle is 50 which the 2 halfs equal 100 and we already know the top side measurement which is 40 and 2 sides of 40 equal 80 and there’s 20 left so this concludes that’s the sides are 10 and square sides are congruent to each other which concludes that’s the AREA is 40
Answer:
C ![11+13+15+17+...](https://tex.z-dn.net/?f=11%2B13%2B15%2B17%2B...)
Step-by-step explanation:
Consider the series
![\sum\limits_{n=3}^{\infty}(2n+5)](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bn%3D3%7D%5E%7B%5Cinfty%7D%282n%2B5%29)
The nth term of series is ![a_n=2n+5](https://tex.z-dn.net/?f=a_n%3D2n%2B5)
The bottom index tells you that n starts changing from 3, so
![a_3=2\cdor 3+5=11\\ \\a_4=2\cdot 4+5=13\\ \\a_5=2\cdot 5+5=15\\ \\a_6=2\cdot 6+5=17\\ \\...](https://tex.z-dn.net/?f=a_3%3D2%5Ccdor%203%2B5%3D11%5C%5C%20%5C%5Ca_4%3D2%5Ccdot%204%2B5%3D13%5C%5C%20%5C%5Ca_5%3D2%5Ccdot%205%2B5%3D15%5C%5C%20%5C%5Ca_6%3D2%5Ccdot%206%2B5%3D17%5C%5C%20%5C%5C...)
Thus, the sum of all terms is
![11+13+15+17+...](https://tex.z-dn.net/?f=11%2B13%2B15%2B17%2B...)
Could you tell me that website? I think I know the answers