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iragen [17]
3 years ago
6

Nate measured an apartment and made a scale drawing. The living room is 2 millimeters long in the drawing. The actual living roo

m is 10 meters long. What scale factor does the drawing use?
Simplify your answer and write it as a ratio, using a colon.
Mathematics
1 answer:
Vlada [557]3 years ago
8 0
Alright! So, we know that in his drawing the living room is 2 millimeters long, and in reality it is 10 meters long. So, to find out how much 1 millimeter is we divide 10 by 2.

[10 ÷ 2] = 5. 

One millimeter is 5 meters in reality.

The ratio is 1:5
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What are the possible numbers of positive, negative, and complex zeros of f(x) = −3x4 −
Anna007 [38]

Answer:

b.

Step-by-step explanation:

We have to look at sign changes in f(x) to determine the possible positive real roots.

f(x)=-3x^4-5x^3-x^2-8x+4

There is only one sign change here, between the -8x and the +4.  So that means there is only 1 possible real positive root.

Now we have to look at sign changes in f(-x) to determine the possible negative real roots.

f(-x)=-3x^4+5x^3-x^2+8x+4

There are 3 sign changes here.  That means there are either 3 negative roots or 3-2 = 1 negative root.  So we have:

1 positive

3 or 1 negative

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If we have 1 positive and 1 negative, we have to have 2 imaginary

If we have 1 positive and 3 negative, we have to have 0 imaginary

Keep in mind that the total number or roots--positive, negative, imaginary--have to add up to equal the degree of the polynomial.  This is a 4th degree polynomial, so we will have 4 roots.

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3 years ago
How many extraneous solutions does the equation below have?
aev [14]

Answer:

A solution is said to be extraneous, if it is a zero of the equation, but it does not satisfy the equation,when substituted in the original equation,L.H.S≠R.H.S. 

The given equation consisting of  variable , m is

   \frac{2 m}{2 m+3} -\frac{2 m}{2 m-3}=1\\\\ 2 m[\frac{1}{2 m+3} -\frac{1}{2 m-3}]=1\\\\ 2 m\times \frac{[2 m-3 -2 m- 3]}{4m^2-9}=1\\\\ -6 \times 2 m=4 m^2 -9\\\\ 4 m^2 +1 2 m -9=0\\\\m=\frac{-12 \pm\sqrt{12^2-4 \times 4 \times (-9)}}{2\times 4}\\\\m=\frac{-12 \pm \sqrt {144+144}}{8}\\\\m=\frac{-12 \pm \sqrt {288}}{8}\\\\m=\frac{-12 \pm 12 \sqrt{2}}{8}\\\\m=\frac{3}{2}\times(-1 \pm \sqrt{2})

None of the two solution

m=\frac{3}{2}\times(-1 +\sqrt{2}) and , m=\frac{3}{2}\times(-1 -\sqrt{2}), is extraneous.

Here, L.H.S= R.H.S

Option A: 0→ extraneous

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