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geniusboy [140]
4 years ago
15

Which statement about this equation is true?

Mathematics
1 answer:
WINSTONCH [101]4 years ago
8 0
The equation has one solution
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Can u help me with my work don’t send a link !!!
lisov135 [29]

Answer:

A and C

Step-by-step explanation:

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Change: f(x) = (x+2) (x-2) to standard form
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X^2 -4

Step-by-step explanation:

Not entirely sure but would u just expand the brackets. As this is a difference between 2 squares the answer should be correct but it not sure.

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in abbys grade, there 70 students. currently, 10% of them are enrolled in health. how many students are enrolled in health?​
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Step-by-step explanation:

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3 years ago
Consider the formula d = , where
Zepler [3.9K]

Answer:

kg/ m³

Step-by-step explanation:

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3 years ago
<img src="https://tex.z-dn.net/?f=x%5Ea%3Dy%5Eb%3Dz%5Ec%20%5C%5Cxyz%3D1%5C%5Cab%2Bbc%2Bca%3D%3F" id="TexFormula1" title="x^a=y^b
SashulF [63]

Answer:

<em>ab+bc+ca=0</em>

Step-by-step explanation:

We are given:

x^a=y^b=z^c

xyz=1

Separating equations:

x^a=z^c

y^b=z^c

Solving the first equation for x:

\displaystyle x=z^\frac{c}{a}

Solving the second equation for y:

\displaystyle y=z^\frac{c}{b}

Substituting in

xyz=1:

\displaystyle z^\frac{c}{a}z^\frac{c}{b}z=1

Adding the exponents:

\displaystyle z^{\frac{c}{a}+\frac{c}{b}+1}=1

Since 1=z^0:

\displaystyle z^{\frac{c}{a}+\frac{c}{b}+1}=z^0

The base is z on both sides so we get rid of them:

\displaystyle \frac{c}{a}+\frac{c}{b}+1=0

Multiplying by ab:

bc+ac+ab=0

Reordering:

ab+bc+ca=0

5 0
3 years ago
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