Answer:
![a_9=48](https://tex.z-dn.net/?f=a_9%3D48)
Step-by-step explanation:
we are given
sequence is geometric
so, we can use nth term formula
![a_n=a_1(r)^{n-1}](https://tex.z-dn.net/?f=a_n%3Da_1%28r%29%5E%7Bn-1%7D)
we have
![a_1=3](https://tex.z-dn.net/?f=a_1%3D3)
![r=\sqrt{2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B2%7D)
we have to find a9
so, we can plug n=9
we get
![a_9=3(\sqrt{2})^{9-1}](https://tex.z-dn.net/?f=a_9%3D3%28%5Csqrt%7B2%7D%29%5E%7B9-1%7D)
![a_9=2^4\cdot \:3](https://tex.z-dn.net/?f=a_9%3D2%5E4%5Ccdot%20%5C%3A3)
![a_9=48](https://tex.z-dn.net/?f=a_9%3D48)
Answer:
Tan C = 3/4
Step-by-step explanation:
Given-
∠ A = 90°, sin C = 3 / 5
<u>METHOD - I</u>
<u><em>Sin² C + Cos² C = 1</em></u>
Cos² C = 1 - Sin² C
Cos² C = ![1 - \frac{9}{25}](https://tex.z-dn.net/?f=1%20-%20%5Cfrac%7B9%7D%7B25%7D)
Cos² C = ![\frac{25 - 9}{25}](https://tex.z-dn.net/?f=%5Cfrac%7B25%20-%209%7D%7B25%7D)
Cos² C = ![\frac{16}{25}](https://tex.z-dn.net/?f=%5Cfrac%7B16%7D%7B25%7D)
Cos C = ![\sqrt{\frac{16}{25} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B16%7D%7B25%7D%20%7D)
Cos C = ![\frac{4}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B5%7D)
As we know that
Tan C = ![\frac{Sin C}{Cos C }](https://tex.z-dn.net/?f=%5Cfrac%7BSin%20C%7D%7BCos%20C%20%7D)
<em>Tan C =
</em>
<em>Tan C =
</em>
<u>METHOD - II</u>
Given Sin C = ![\frac{3}{5} = \frac{Height}{Hypotenuse}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B5%7D%20%3D%20%5Cfrac%7BHeight%7D%7BHypotenuse%7D)
therefore,
AB ( Height ) = 3; BC ( Hypotenuse) = 5
<em>∵ ΔABC is Right triangle.</em>
<em>∴ By Pythagorean Theorem-</em>
<em>AB² + AC² = BC²</em>
<em>AC² </em><em>= </em><em>BC² </em><em>- </em><em> AB</em><em>² </em>
<em>AC² = 5² - 3²</em>
<em>AC² = 25 - 9</em>
<em>AC² = 16</em>
<em>AC ( Base) = 4</em>
<em>Since, </em>
<em>Tan C =
</em>
<em>Tan C =
</em>
<em>Hence Tan C =
</em>
<em />
Any fraction that does not equal 1/2.
Answer: B. 3x + 1/5
tom's pencil is longer than Ellen's pencil:
5x + 2/5 - (2x + 1/5) = 5x - 2x + 2/5 - 1/5 = 3x + 1/5 (cm)
Step-by-step explanation: