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NISA [10]
3 years ago
11

How can you test to prove that one of your circuits is a series circuit?

Mathematics
1 answer:
choli [55]3 years ago
7 0
If you move one component ( like a light) then no current will pass when you activate the switch.

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What is the answer please help no links no links please help
schepotkina [342]
180cm. The square is 6 x 6 = 36 and the triangles = 12 x 6 = 72 and then divide it by 2 = 36 multiply that by 5 and you get 180cm
7 0
3 years ago
Read 2 more answers
Please help!!!!!!!!! (Solve The quadratic equation by completing the square) 2x^2+12x=66 fill in the value of a and b to complet
yKpoI14uk [10]

I hope this it

Explanation:

5 0
2 years ago
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If a and b are independent events with p(a) = 0.05 and p(b) = 0.65, then the conditional probability p(a |
alukav5142 [94]
Two events are said to be independent when the occurence of one event does not in any way affects the occurence of the other event.

For independent events, the probablilty of event A happening given that event B has happened is given by the probability of event A happening.

i.e. P(a | b) = P(a).

Therefore, given that p(a) = 0.05 and P(b) = 0.65, then p(a | b) = p(a) = 0.05.
4 0
3 years ago
Answer the question in the picture
nekit [7.7K]

Recall the angle sum identities:

\sin(x+y)=\sin x\cos y+\cos x\sin y

\cos(x+y)=\cos x\cos y-\sin x\sin y

Now,

\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}

Divide through numerator and denominator by \cos x\cos y to get

\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}

Next, we use the fact that x,y lie in the first quadrant to determine that

\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2

\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}

So we then have

\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}

\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1

Finally,

\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73

4 0
3 years ago
If AB = 10, BD = 5, and DE= 12, what is the length of BC?
FrozenT [24]

Answer:

BC = 24

Step-by-step explanation:

In the picture attached, the missing triangles are shown:

<u>Data</u>

  • AB = 10
  • BD = 5
  • DE= 12
  • Then, AD = 10-5 = 5

As a consequence of the "Side Splitter" Theorem:

AD/DE = AB/BC

Replacing with data and solving for BC

5/12 = 10/BC

BC = 10*12/5 = 24

4 0
3 years ago
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