Solving equation 2 / N + 3 / N + 1 + 4 / N + 2 = 133 / 60

9/n+3=133/60;
9/n=-47/60;
n=-540/47

Calculation: sin17 ° cos43 ° + cos17 ° sin43 °

The original formula is sin (17 ° + 43 °) = sin 60 ° = 32

How to solve the equation (30% x + 38): x = 4:7

(30%x+38):x=4:7
4x=7*(0.3x+38)
4x=2.1x+266
4x-2.1x=266
1.9x=266
x=266/1.9
x=140

Three quarters of a meter is equal to a fraction of three meters

Three quarters of a meter is equal to one fourth of a meter

If the exponential equation 9 ^ x + 3 ^ x + a = 0 has a solution, find the value of A
RT

The two answers above are all wrong. It can be verified that a = 1 / 4,3 ^ x = - 1 / 2 has no solution
t=3^x>0,f(x)=t²+t+a
Therefore, if △ = 1-4a ≥ 0 and there is at least one positive root, the observed image has f (0)

Using Tan α to express trigonometric 1 / (COS ^ 2 α sin ^ 2 α)

1/（cos^2αsin^2α）=（cos^2α+sin^2α)^2/（cos^2αsin^2α）=[(cosa)^4+2(sina)^2(cosa)^2+(cosa)^4]/[(cosa)^2(sina)^2]=[(cosa)^4+(cosa)^4]/[(cosa)^2(sina)^2]+2=[1+(tana)^4]/(tana)^2+2

Find the next number

8+8 = 4^2
8+56 = 8^2
∴56+x = 8^2
∴x = 8
That is, the next number is 8

To prove Tana + 1 / Tan (π / 4 + A / 2) = 1 / cosa

Tan (π / 4 + A / 2) = (1 + Tan (A / 2)) / (1-tan (A / 2)) repeatedly uses the universal formula Tana + 1 / Tan (π / 4 + A / 2) = 2tan (α / 2) / [1-tan ^ 2 (α / 2)] + (1-tan (A / 2)) / (1 + Tan (A / 2)) = [1 + Tan ^ 2 (α / 2)] / [1-tan ^ 2 (α / 2)] = 1 / cosa

Given p (4, - 9), q (- 2,3), and the intersection of Y-axis and segment PQ and m, if MQ = xqp, find the value of X

Let y = KX, B substitute (4, - 9) (- 2,3) into k = 2, B = - 1, when y = 0, m (- 1,0) x = 2

On the proof of the periodicity of a function
It is proved that the period of F (1 + x) = f (1-x) is 1, which is determined by the definition of period

Certification:
Let x + 1 = t
therefore
x=t-1
that is
1-x=2-t
therefore
f(1+x)=f(1-x)
It can be written as
F (T) = f (2-T) but I can't go on with your question
The condition is about even function, right?

Solving equation 2 / N + 3 / N + 1 + 4 / N + 2 = 133 / 60