Find the nth term of a gp with (a. find term = 5, common ratio = 2

Mathematics

1 answer:

Rina8888 [55]2 years ago

7 0

Step-by-step explanation:

we know,

$tn= a {(r)}^{n - 1} \\ T5 = a {(2)}^{5 - 1} \\ t5 = a \times {2}^{4} \\ = 16a$

Therefore, nth term of a given gp is

16a

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in a classroom 5/8 of the students are wearing blue shirts and 1/4 for wearing white shirts there are 24 students in the classro

Oliga [24]

Answer:

Step-by-step explanation:

Those wearing a shirt of another color are ...

1 - 5/8 -1/4 = 8/8 -5/8 -2/8 = 1/8

of the total number of students in the classroom

(1/8)×(24 students) = 3 students . . . . wearing another color

_____

<em>Alternate solution</em>

With the given information, you know ...

(5/8)(24) = 15 . . . students wear blue

(1/4)(24) = 6 . . . . students wear white

24 -15 -6 = 3 . . . students wear another color

6 0

2 years ago

If sin A = 3/8, find the value of cosec A - sec A.

alexira [117]

Answer:

$\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$

Step by step explanation:

$\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies 1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\$

$\implies \dfrac 1{\cos A} = \pm\dfrac{8}{\sqrt{55}}$

$\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$