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ICE Princess25 [194]

3 years ago

14

Help me out brothas

1 answer:

Jobisdone [24]3 years ago

4 0

$C = \frac{5}{9} (f - 32)$

$c=\frac{5(f-32)}{9}$

c×9=5(f−32)

9c=5(f−32)

$\frac{9c}{5}=f-32$

$\frac{9c}{5}+32=f$

$f=\frac{9c}{5}+32$

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2 years ago

Please help... I have no clue

muminat

Answer:

OPTION C: Sin C - Cos C = s - r

Step-by-step explanation:

ABC is a right angled triangle. ∠A = 90°, from the figure.

Therefore, BC = hypotenuse, say h

Now, we find the length of AB and AC.

We know that: $\textbf{Sin A} = \frac{\textbf{opp}}{\textbf{hyp}}$

and $\textbf{Cos A} = \frac{\textbf{adj}}{\textbf{hyp}}$

Given, Sin B = r and Cos B = s

⇒ $Sin B = r = \frac{opp}{hyp} = \frac{AC}{BC} = \frac{AC}{h}$

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Hence, the length of the side AC = rh

Now, to compute the length of AB, we use Cos B.

$Cos B = s = \frac{adj}{hyp} = \frac{AB}{BC} = \frac{AB}{h}$

⇒ $\textbf{AB} = \textbf{sh}$

Hence, the length of the side AB = sh

Now, we are asked to compute Sin C - Cos C.

$Sin C = \frac{opp}{hyp}$

⇒ $Sin C = \frac{AB}{BC}$

$= \frac{sh}{h}$

= s

Sin C = s

$Cos C = \frac{adj}{hyp}$

$\implies Cos C = \frac{AC}{BC}$

⇒ Cos C = $\frac{rh}{h}$

Therefore, Cos C = r

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3 years ago

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Step-by-step explanation:

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2 years ago

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Katyanochek1 [597]

<h2>Answer:</h2>

$\boxed{R_{4}=20\Omega}$

<h2>Step-by-step explanation:</h2>

If resistors are in series, so the current $I$ is the same in all of them. In this problem we have four resistors. So, we can get a relationship between the Equivalent resistance of series combination and the four resistors as follows:

$R_{eq}=R_{1}+R_{2}+R_{3}+R_{4}$

$R_{eq}$ is the total resistance $50\Omega$ . Moreover:

$R_{1}=5\Omega \\ R_{2}=10\Omega \\ R_{3}=15\Omega$

Therefore:

$50=5+10+15+R_{4} \\ \\ R_{4}=50-(5+10+15) \\ \\ R_{4}=50-30 \\ \\ \boxed{R_{4}=20\Omega}$

7 0

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4 0

3 years ago