1 + tan ² Ф=sec²Ф
1+(12/5)²=sec²Ф
169/25=sec² Ф
sec Ф=⁺₋√(169/25)=⁺₋13/5
sec Ф=1/cos Ф ⇒cosФ=1/sec Ф
cos Ф>0 ⇔ sec Ф>0 ⇔ sec Ф=+ 13/5
cos Ф=1/secФ
cos Ф=1 / 13/5=5/13
we can calculate the sin Ф, with this method.
sin²Ф + cos²Ф=1 ⇒ sin Ф=⁺₋√(1-cos² Ф)
sin Ф=⁺₋√[1-(5/13)²]=⁺₋12/13
like cos Ф>0 and tan Ф>0 ⇒ sin Ф>0 ⇒sin Ф=12/13
answer: d.12/13
other method
tan Ф=sin Ф / cos Ф
12/5=sin Ф / 5/13
sin Ф=(12/5)*(5/13)=12/13
answer: d.12/13
Solution:
Given that the point P lies 1/3 along the segment RS as shown below:
To find the y coordinate of the point P, since the point P lies on 1/3 along the segment RS, we have

Using the section formula expressed as
![[\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bmx_2%2Bnx_1%7D%7Bm%2Bn%7D%2C%5Cfrac%7Bmy_2%2Bny_1%7D%7Bm%2Bn%7D%5D)
In this case,

where

Thus, by substitution, we have
![\begin{gathered} [\frac{1(2)+2(-7)}{1+2},\frac{1(4)+2(-2)}{1+2}] \\ \Rightarrow[\frac{2-14}{3},\frac{4-4}{3}] \\ =[-4,\text{ 0\rbrack} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5B%5Cfrac%7B1%282%29%2B2%28-7%29%7D%7B1%2B2%7D%2C%5Cfrac%7B1%284%29%2B2%28-2%29%7D%7B1%2B2%7D%5D%20%5C%5C%20%5CRightarrow%5B%5Cfrac%7B2-14%7D%7B3%7D%2C%5Cfrac%7B4-4%7D%7B3%7D%5D%20%5C%5C%20%3D%5B-4%2C%5Ctext%7B%200%5Crbrack%7D%20%5Cend%7Bgathered%7D)
Hence, the y-coordinate of the point P is
Answer:
3^4 / 8^2 or 81/64
Step-by-step explanation:
(-8)^-2 / 3^-4 =
= 3^4 / (-8)^2
= 3^4 / 8^2
= 81/64
Step-by-step explanation:
1 Remove parentheses.
8{y}^{2}\times -3{x}^{2}{y}^{2}\times \frac{2}{3}x{y}^{4}
8y
2
×−3x
2
y
2
×
3
2
xy
4
2 Use this rule: \frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}
b
a
×
d
c
=
bd
ac
.
\frac{8{y}^{2}\times -3{x}^{2}{y}^{2}\times 2x{y}^{4}}{3}
3
8y
2
×−3x
2
y
2
×2xy
4
3 Take out the constants.
\frac{(8\times -3\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}
3
(8×−3×2)y
2
y
2
y
4
x
2
x
4 Simplify 8\times -38×−3 to -24−24.
\frac{(-24\times 2){y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}
3
(−24×2)y
2
y
2
y
4
x
2
x
5 Simplify -24\times 2−24×2 to -48−48.
\frac{-48{y}^{2}{y}^{2}{y}^{4}{x}^{2}x}{3}
3
−48y
2
y
2
y
4
x
2
x
6 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x
a
x
b
=x
a+b
.
\frac{-48{y}^{2+2+4}{x}^{2+1}}{3}
3
−48y
2+2+4
x
2+1
7 Simplify 2+22+2 to 44.
\frac{-48{y}^{4+4}{x}^{2+1}}{3}
3
−48y
4+4
x
2+1
8 Simplify 4+44+4 to 88.
\frac{-48{y}^{8}{x}^{2+1}}{3}
3
−48y
8
x
2+1
9 Simplify 2+12+1 to 33.
\frac{-48{y}^{8}{x}^{3}}{3}
3
−48y
8
x
3
10 Move the negative sign to the left.
-\frac{48{y}^{8}{x}^{3}}{3}
−
3
48y
8
x
3
11 Simplify \frac{48{y}^{8}{x}^{3}}{3}
3
48y
8
x
3
to 16{y}^{8}{x}^{3}16y
8
x
3
.
-16{y}^{8}{x}^{3}
−16y
8
x
3
Done
Answer:
LCM of 6 and 15 by Listing Multiples
Step 2: The common multiples from the multiples of 6 and 15 are 30, 60, . . . Step 3: The smallest common multiple of 6 and 15 is 30.