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3 years ago

How do you write in words 4.110 and 4.10?

Mathematics

2 answers:

ohaa [14]3 years ago

8 0

The <em>correct answers</em> are:

"Four and 11 hundredths" or "four and one hundred ten thousandths"; and

"Four and ten hundredths" or "four and one tenth."

Explanation:

In the first decimal, the digit 4 is in front of the decimal. This is read "Four and". After the decimal, we have 110. We look at the last place, digit, 0. Going by our place value, this is in the thousandths place. We read the block of digits 110 as "one hundred ten" and it is "thousandths." This makes the number "four and one hundred ten thousandths."

Alternatively, if we drop the zero, we would have 4.11. It is still "four and"; this time the last digit is 1, and it is in the hundredths place. 11 is read as "eleven", so this would be "four and eleven hundredths."

For the second number, 4.10, we have a 4 in front of the decimal, so we again have "four and". Our last digit is 0, and it is in the hundredths place; 10 is read "ten", so we have "four and ten hundredths."

Alternatively, if we drop the zero, we have 4.1. This is still "four and"; the 1 is now our last digit, and it is in the tenths place, so we have "four and one tenth."

Lana71 [14]3 years ago

8 0

For this case we have the following numbers:

4.110

4.10

We observed that:

4.110> 4.10

Therefore, we should not confuse both numbers.

The correct way to write both numbers is given by:

4.110 = four point one hundred ten thousandths

4.10 = four point ten hundredths

Answer:

4.110 = four point one hundred ten thousandths

4.10 = four point ten hundredths

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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

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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

$\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23$

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:

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

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

$\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12$

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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