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3 years ago

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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Please help!

Gemiola [76]

Answer:

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4 0

3 years ago

What’s the volume of the cone if the height is 7 cm and the base it 10 cm

pogonyaev

Answer:

23.3333333333...

Step-by-step explanation:

10 x 7 = 70

70/3 = 23.333333333...

5 0

3 years ago

Please take a look at the attachment.

Tanya [424]

Answer:

......

30%..................

3 0

2 years ago

Read 2 more answers

Which of the following are the zeros of the function y= 3(x-a)(x+b) in terms of a and b?

frosja888 [35]

The correct answer is option 2.

3 = 0

x - a = 0

x = a

x + b = 0

x = -b

We do not multiply any of these values by 3 because in a way, 3 is acting as another "zero" of the equation. Though, as shown above, it does not equal 0, it still does not apply to any of the other equations solved above.

Hope this helps!! :)

7 0

3 years ago

Triangle RST has vertices located at R (2, 3), S (4, 4), and T (5, 0).

LiRa [457]

The length of the sides are 2.24, 4.24 and 4.12 units while the slope are 0.5, -1 and -4.

<h3>Linear equation</h3>

A linear equation is in the form:

y = mx + b

where y, x are variables, m is the rate of change and b is the initial value of y.

For RS:

$Length = \sqrt{(y_2-y_1)^2+(x_2-x_1)^2}=\sqrt{(4-3)^2+(4-2)^2}=2.24 \\
\\
slope=\frac{y_2-y_1}{x_2-x_1} =\frac{4-3}{4-2}=0.5$

For RT:

$Length = \sqrt{(y_2-y_1)^2+(x_2-x_1)^2}=\sqrt{(0-3)^2+(5-2)^2}=4.24 \\
\\
slope=\frac{y_2-y_1}{x_2-x_1} =\frac{0-3}{5-2}=-1$

For ST:

$Length = \sqrt{(y_2-y_1)^2+(x_2-x_1)^2}=\sqrt{(0-4)^2+(5-4)^2}=4.12 \\
\\
slope=\frac{y_2-y_1}{x_2-x_1} =\frac{0-4}{5-4}=-4$

The length of the sides are 2.24, 4.24 and 4.12 units while the slope are 0.5, -1 and -4.

Find out more on linear equation at: brainly.com/question/14323743

5 0

2 years ago