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emmainna [20.7K]
1 year ago
12

Im so confused pls help

Mathematics
1 answer:
LuckyWell [14K]1 year ago
6 0

Answer:

  (d)  f(x) = log₆(x)

Step-by-step explanation:

If we use y in place of f(x), we see that ...

  x = 6^y

Taking logs of both sides, we get ...

  log(x) = y·log(6)

  y = log(x)/log(6)

  y = log₆(x) . . . . . . . using the change of base formula

  f(x) = log₆(x)

_____

Or you can get there more directly using the relation between logs and exponentials:

  \displaystyle a = b^c\  \leftrightarrow\ c=\log_b(a)

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The volume of a rectangular prism is represented by the function x3 + 11x2 + 20x – 32. The width of the box is x – 1 while the h
Oksi-84 [34.3K]

we know that

the volume of a rectangular prism is equal to

V=L*W*H

where

L is the length of the box

W is the width of the box

H is is the height of the box

in this problem we have

V=x^{3\ }+11x^2+20x-32

W=x-1

H=x+8

L=?

<u>Find the length of the box</u>

Using a graph tool-------> we will determine the roots of the equation of volume

see the attached figure

the roots are

x=-8\\x=-4\\ x=1

so

x^{3\ }+11x^2+20x-32=(x+8)*(x+4)*(x-1)

therefore

<u>the answer is</u>

the length of the box is equal to (x+4)

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17. Evaluate each pair of expressions.<br> a. (-3)-8 and -3-8<br> b. (-3)-9 and -3-9
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A store is having a sale on chocolate chips and walnuts. For 3 pounds of chocolate chips and 5 pounds of walnuts, the total cost
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\bf \stackrel{\textit{3 lbs of "c"}}{3c}+\stackrel{\textit{5 lbs of "w"}}{5w}~~=~~\stackrel{\textit{costs}}{15} \\\\\\ \stackrel{\textit{12 lbs of "c"}}{12c}+\stackrel{\textit{2 lbs of "w"}}{2w}~~=~~\stackrel{\textit{costs}}{33} \end{cases}\qquad \impliedby \textit{let's use elimination} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{llccccccl} 3c+5w=15&\times (-4)\implies &-12c&+&-20w&=&-60\\ 12c+2w=33&&12c&+&2w&=&33\\ \cline{3-7}\\ &&0&&-18w&=&-27 \end{array}

\bf -18w=-27\implies w=\cfrac{-27}{-18}\implies \blacktriangleright w=\cfrac{3}{2} \blacktriangleleft \\\\\\ \stackrel{\textit{substituting on the 1st equation}}{3c+5\left(\cfrac{3}{2} \right)=15}\implies 3c+\cfrac{15}{2}=15 \implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2\left( 3c+\cfrac{15}{2} \right)=2(15)} \\\\\\ 6c+15=30\implies 6c=15\implies c=\cfrac{15}{6}\implies \blacktriangleright c=\cfrac{5}{2} \blacktriangleleft

6 0
3 years ago
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