Answer:
48
24
Step-by-step explanation:
16x3=48
16/2=8
8x3=24
From Pythagorean Identities <span><span><span>sin^2</span>x</span>+<span><span>cos^2</span>x</span>=1</span>
and from that identities we can have <span><span>sinx</span>+<span>cosx</span>=1</span>, taking <span><span>sinx</span>=1−<span>cosx</span></span> and substitute in the equation
<span><span>sinx</span>+<span>cosx</span>=1</span>
<span><span>(1−<span>cosx</span>)</span>+<span>cosx</span>=1</span>
Remove the parenthesis
<span>1−<span>cosx</span>+<span>cosx</span>=1</span>
cancel <span>cosx</span>
<span>1=1</span>
sqrt (3) * sqrt(16) = 4sqrt(3) = irrational
2sqrt(9) *sqrt(4) = 6*2 = 12 rational
7 sqrt(3)* sqrt(3) = 7*3 =21 rational
sqrt(5) * sqrt(5) = 5 rational
Answer:
![\sec y=\dfrac{q}{r}](https://tex.z-dn.net/?f=%5Csec%20y%3D%5Cdfrac%7Bq%7D%7Br%7D)
Step-by-step explanation:
Given that,
![\sin y=\dfrac{7}{q}](https://tex.z-dn.net/?f=%5Csin%20y%3D%5Cdfrac%7B7%7D%7Bq%7D)
and
![\tan y=\dfrac{7}{r}](https://tex.z-dn.net/?f=%5Ctan%20y%3D%5Cdfrac%7B7%7D%7Br%7D)
We need to find the value of
. We know that,
![\sec\theta=\dfrac{1}{\cos\theta}](https://tex.z-dn.net/?f=%5Csec%5Ctheta%3D%5Cdfrac%7B1%7D%7B%5Ccos%5Ctheta%7D)
Also,
![\tan\theta=\dfrac{\sin\theta}{\cos\theta}\\\\\cos\theta=\dfrac{\sin\theta}{\tan\theta}](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cdfrac%7B%5Csin%5Ctheta%7D%7B%5Ccos%5Ctheta%7D%5C%5C%5C%5C%5Ccos%5Ctheta%3D%5Cdfrac%7B%5Csin%5Ctheta%7D%7B%5Ctan%5Ctheta%7D)
Substitute all the values,
![\cos y=\dfrac{\dfrac{7}{q}}{\dfrac{7}{r}}\\\\=\dfrac{7}{q}\times \dfrac{r}{7}\\\\\cos y=\dfrac{r}{q}](https://tex.z-dn.net/?f=%5Ccos%20y%3D%5Cdfrac%7B%5Cdfrac%7B7%7D%7Bq%7D%7D%7B%5Cdfrac%7B7%7D%7Br%7D%7D%5C%5C%5C%5C%3D%5Cdfrac%7B7%7D%7Bq%7D%5Ctimes%20%5Cdfrac%7Br%7D%7B7%7D%5C%5C%5C%5C%5Ccos%20y%3D%5Cdfrac%7Br%7D%7Bq%7D)
So,
![\sec y=\dfrac{1}{\dfrac{r}{q}}\\\\\sec y=\dfrac{q}{r}](https://tex.z-dn.net/?f=%5Csec%20y%3D%5Cdfrac%7B1%7D%7B%5Cdfrac%7Br%7D%7Bq%7D%7D%5C%5C%5C%5C%5Csec%20y%3D%5Cdfrac%7Bq%7D%7Br%7D)
So, the correct option is (a) i.e.
.
Given 2 points A(a,b) and M(k,l), the distance between them is found by the formula:
![|AM|= \sqrt{ (a-k)^{2} + (b-l)^{2} }](https://tex.z-dn.net/?f=%7CAM%7C%3D%20%5Csqrt%7B%20%28a-k%29%5E%7B2%7D%20%2B%20%28b-l%29%5E%7B2%7D%20%7D%20)
Let (k, l)=(2, 3), substituting in the above formula:
![12= \sqrt{ (a-2)^{2} + (b-3)^{2} }](https://tex.z-dn.net/?f=12%3D%20%5Csqrt%7B%20%28a-2%29%5E%7B2%7D%20%2B%20%28b-3%29%5E%7B2%7D%20%7D)
![(a-2)^{2} + (b-3)^{2}=12 ^{2}=144](https://tex.z-dn.net/?f=%28a-2%29%5E%7B2%7D%20%2B%20%28b-3%29%5E%7B2%7D%3D12%20%5E%7B2%7D%3D144%20)
check each of the pairs given:
for (a, b)=(14, 15):
![(a-2)^{2} + (b-3)^{2}=(14-2)^{2} + (15-3)^{2}=(12)^{2} + (12)^{2}=288](https://tex.z-dn.net/?f=%28a-2%29%5E%7B2%7D%20%2B%20%28b-3%29%5E%7B2%7D%3D%2814-2%29%5E%7B2%7D%20%2B%20%2815-3%29%5E%7B2%7D%3D%2812%29%5E%7B2%7D%20%2B%20%2812%29%5E%7B2%7D%3D288)
for (a, b)=(2, -9):
![(2-2)^{2} + (-9-3)^{2}=0+(-12)^{2}=144](https://tex.z-dn.net/?f=%282-2%29%5E%7B2%7D%20%2B%20%28-9-3%29%5E%7B2%7D%3D0%2B%28-12%29%5E%7B2%7D%3D144)
for (a, b)=(-2, -3):
![(-2-2)^{2} + (-3-3)^{2}=(-4)^{2} + (-9)^{2}=16+81=97](https://tex.z-dn.net/?f=%28-2-2%29%5E%7B2%7D%20%2B%20%28-3-3%29%5E%7B2%7D%3D%28-4%29%5E%7B2%7D%20%2B%20%28-9%29%5E%7B2%7D%3D16%2B81%3D97)
for (a, b)=(-9, 2):
![(-9-2)^{2} + (2-3)^{2}=(-11)^{2} + (-1)^{2}=121+1=122](https://tex.z-dn.net/?f=%28-9-2%29%5E%7B2%7D%20%2B%20%282-3%29%5E%7B2%7D%3D%28-11%29%5E%7B2%7D%20%2B%20%28-1%29%5E%7B2%7D%3D121%2B1%3D122)
Answer: (2, -9)