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Finger [1]
3 years ago
12

A laboratory technician needs to make a ​81-liter batch of a ​20% acid solution. How can the laboratory technician combine a bat

ch of an acid solution that is pure acid with another that is ​10% to get the desired​ concentration?
Mathematics
1 answer:
Ludmilka [50]3 years ago
5 0
Honedtlrfjdjdjdjfjfjjfjfjfjjf
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Which of the following is an identity? A. sin2x sec2x + 1 = tan2x csc2x B. sin2x - cos2x = 1 C. (cscx + cotx)2 = 1 D. csc2x + co
Ne4ueva [31]
There are three 'Pythagorean' identities that we can look at and they are

sin²(x) + cos²(x) = 1
tan²(x) + 1 = sec²(x) 
1 + cot²(x) = csc²(x)

We can start by checking each option to see which one would give us any of the 'Pythagorean' identities as its simplest form

Option A:

sin²(x) sec²(x) + 1 = tan²(x) csc²(x)

Rewriting sec²(x) as 1/cos²(x)
Rewriting tan²(x) as sin²(x)/cos²(x)
Rewriting csc²(x) as 1/sin²(x)

We have

sin^{2}(x)[ \frac{1}{ cos^{2}(x) }]+1=[ \frac{ sin^{2}( x)}{ cos^{2} (x)}][ \frac{1}{ sin^{2}(x) } ]
[\frac{ sin^{2}(x) }{ cos^{2}(x) } ]+1= \frac{1}{ cos^{2}(x) }
tan^{2}(x)+1= sec^{2}(x)

Option B:

sin²(x) - cos²(x) = 1

This expression is already in the simplest form, cannot be simplified further

Option C:

[ csc(x) + cot(x) ]² = 1

Rewriting csc(x) as 1/sin(x)
Rewriting cot(x) as cos(x)/sin(x)

We have

[ \frac{1}{sin(x)}+ \frac{cos(x)}{sin(x)}] ^{2} =1
\frac{1}{sin^2(x)}+2( \frac{1}{sin(x)})( \frac{cos(x)}{sin(x)})+ \frac{cos^2(x)}{sin^2(x)}=1csc^2(x)+2csc^2(x)cos(x)+cot^2(x)=1

Option D:

csc²(x) + cot²(x) = 1

Rewriting csc²(x) as 1/sin²(x) and cot²(x) as cos²(x)/sin²(x)

\frac{1}{sin^2(x)}+ \frac{cos^2(x)}{sin^2(x)}=1
\frac{1+cos^2(x)}{sin^2(x)} =1
1+cos^2(x)=sin^2(x)
1=sin^2(x)-cos^2(x)

from our working out we can see that option A simplified into one of 'Pythagorean' identities, hence the correct answer
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When solving for the value of x in the equation 4(x - 1) + 3 = 18,
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(1) distributive is the answer. Distributive property just means 4(x-1)=4x-4
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