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shepuryov [24]
3 years ago
10

Heba ate 1/12 of a box of cereal. Now the box is 3/4 full.

Mathematics
2 answers:
ivanzaharov [21]3 years ago
8 0
I believe there was 5/6s of the box there before.
diamong [38]3 years ago
7 0

Answer:

The fraction of a full box was \frac{5}{6}  before haba ate.

Step-by-step explanation:

Consider the provided information.

Heba ate 1/12 of a box of cereal. Now the box is 3/4 full.

\frac{
3}{4}=\frac{3}{4}\times 1

The above equation can be written as:

\frac{
3}{4}=\frac{3}{4}\times{\frac{3}{3}}

\frac{
3}{4}=\frac{9}{12}

Thus, the Box is still 9/12 full.

As haba ate 1/12 of it simply add 1/12 and 9/12 as shown:

\frac{9}{12}+\frac{1}{12}

\frac{9+1}{12}

\frac{10}{12}

\frac{5}{6}

Hence, the fraction of a full box was \frac{5}{6}  before haba ate.

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Answer:

The graph of g is the graph of f shifted up by 3 units.

Step-by-step explanation:

Consider the graph of a function r with real numbers k and h.

Transformation Effect

r(x) + k shifts the graph up k units

r(x) - k shifts the graph down k units

r(x + h) shifts the graph to the left h units

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It is given that g(x) = f(x) + 3. Therefore, the graph of g is the graph of f shifted up by 3 units.

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3 years ago
In the expression (44 ÷ 4 + 10) - 20 , with which operation should we start?
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4 0
2 years ago
Read 2 more answers
If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)
Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

\log_dabc=\log_da+\log_db+\log_dc

Then using the change of base formula, we can derive the relationship

\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}

This immediately tells us that

\log_dc=\dfrac1{\log_cd}=\dfrac12

Notice that none of a,b,c,d can be equal to 1. This is because

\log_1x=y\implies1^{\log_1x}=1^y\implies x=1

for any choice of y. This means we can safely do the following without worrying about division by 0.

\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}

so that

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Similarly,

\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}

so that

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So we end up with

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###

Another way to do this:

\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}

\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}

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Then

abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}

So we have

\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}

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slavikrds [6]

Answer:

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Step-by-step explanation:

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