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3 years ago

Find the value of h rounded to the nearest tenth

Mathematics

2 answers:

Marrrta [24]3 years ago

4 0

Answer:

h = 6 .7 units.

Step-by-step explanation:

Given : Triangle .

To find : Find the value of h rounded to the nearest tenth.

Solution : We have given that

Triangle with hypotenuses = h .

Angle = 48 degree .

Opposite side = 5 .

By the trigonometric ratio : $sin(angle) =\frac{opposite}{hypotenuse}$ .

$sin(48) =\frac{5}{h}$ .

0.743 = $\frac{5}{h}$ .

On multiplying both sides by h.

0.743 * h = 5

On dividing by 0 .743

h = $\frac{5}{0.743}$

h = 6.72

Therefore, h = 6 .7 units.

3241004551 [841]3 years ago

3 0

For angle G, FH is the opposite leg.
h is the hypotenuse.

The trig ratio that relates the opposite leg and the hypotenuse is the sine.

$\sin \theta = \dfrac{opp}{hyp}$

$\sin 48^\circ = \dfrac{5}{h}$

$h = \dfrac{5}{\sin 48^\circ}$

$h = 6.7$

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The sum of the first n terms of a geometric series is 364? The sum of their reciprocals 364/243. If the first term is 1, find n

Afina-wow [57]

If the geometric series has first term $a$ and common ratio $r$ , then its $N$ -th partial sum is

$\displaystyle S_N = \sum_{n=1}^N ar^{n-1} = a + ar + ar^2 + \cdots + ar^{N-1}$

Multiply both sides by $r$ , then subtract $rS_N$ from $S_N$ to eliminate all the middle terms and solve for $S_N$ :

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The $N$ -th partial sum for the series of reciprocal terms (denoted by $S'_N$ ) can be computed similarly:

$\displaystyle S'_N = \sum_{n=1}^N \frac1{ar^{N-1}} = \frac1a + \frac1{ar} + \frac1{ar^2} + \cdots + \frac1{ar^{N-1}}$

$\dfrac{S'_N}r = \dfrac1{ar} + \dfrac1{ar^2} + \dfrac1{ar^3} + \cdots + \dfrac1{ar^N}$

$\implies \left(1 - \dfrac1r\right) S'_N = \dfrac1a - \dfrac1{ar^N}$

$\implies S'_N = \dfrac{1 - \frac1{r^N}}{a\left(1 - \frac1r\right)} = \dfrac{r^N - 1}{a(r^N - r^{N-1})} = \dfrac{1 - r^N}{a r^{N-1} (1 - r)}$

We're given that $a=1$ , and the sum of the first $n$ terms of the series is

$S_n = \dfrac{1-r^n}{1-r} = 364$

and the sum of their reciprocals is

$S'_n = \dfrac{1 - r^n}{r^{n-1}(1 - r)} = \dfrac{364}{243}$

By substitution,

$\dfrac{1 - r^n}{r^{n-1}(1-r)} = \dfrac{364}{r^{n-1}} = \dfrac{364}{243} \implies r^{n-1} = 243$

Manipulating the $S_n$ equation gives

$\dfrac{1 - r^n}{1-r} = 364 \implies r (364 - r^{n-1}) = 363$

so that substituting again yields

$r (364 - 243) = 363 \implies 121r = 363 \implies \boxed{r=3}$

and it follows that

$r^{n-1} = 243 \implies 3^{n-1} = 3^5 \implies n-1 = 5 \implies \boxed{n=6}$

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You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

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