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kondor19780726 [428]
3 years ago
15

a secant and tangent are drawn to a circle from a common exterior point. If the length is 14 inches and the length of the extern

al segment of the secant is 4 inches, then it which of the following is the length,in inches, of the internal segment of the secant?

Mathematics
1 answer:
hoa [83]3 years ago
5 0

Answer:

Correct option: (3) 45

Step-by-step explanation:

Let's call the length of the tangent 'T', the length of the external segment 'E' and the length of the internal segment 'A'.

If we have a secant and a tangent to a circle, we can use the following property:

T^2 = E * (A + E)

If the tangent is 14 inches and the external segment is 4 inches, we have that:

14^2 = 4 * (A + 4)

196 = 4A + 16

4A = 180

A = 45

Correct option: (3)

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Step-by-step explanation:

To find the equation of the line, we want to find the slope with the formula m=\frac{y_2-y_1}{x_2-x_1}.

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Now that we know th eslope, we can start to fill out the equation. To find the y-intercept, we can plug in a given point and solve.

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If log2 5 = k, determine an expression for log32 5 in terms of k.
lukranit [14]

Answer:

log_3_2(5)=\frac{1}{5} k

Step-by-step explanation:

Let's start by using change of base property:

log_b(x)=\frac{log_a(x)}{log_a(b)}

So, for log_2(5)

log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)

Now, using change of base for log_3_2(5)

log_3_2(5)=\frac{log(5)}{log(32)}

You can express 32 as:

2^5

Using reduction of power property:

log_z(x^y)=ylog_z(x)

log(32)=log(2^5)=5log(2)

Therefore:

log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)

As you can see the only difference between (1) and (2) is the coefficient \frac{1}{5} :

So:

\frac{log(5)}{log(2)} =k\\

log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k

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Answer:

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Answer:

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Since we are not given a zero on the number line, we do the following:

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Next we find how many units away are each number from the middle:

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Let's use the formula:

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x = 7 or x = 3

LOOK AT THE GRAPH AND NUMBER LINE

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