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3 years ago

Prove that the square of an odd number is always 1 more than a multiple of 4

Mathematics

1 answer:

loris [4]3 years ago

5 0

Answer:

By these examples you are able to see that the square of an odd number is always 1 more than a multiple of 4.

Step-by-step explanation:

For examples,

Let's consider squares of 3, 11, 25, 37 and 131.

${3}^{2} = 9$

8 is a multiple of 4, and 9 is more than 8.

${11}^{2} = 121$

120 is a multiple of 4 and 121 is one more than it.

${25}^{2} = 625$

624 is a multiple of 4 and 625 is one more than it.

${37}^{2} = 1369$

1368 is a multiple of 4 and 1369 is one more than 1368.

${131}^{2} = 17161$

17160 is a multiple of 4.

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What two numbers multiply to -50 and add up to -23

umka2103 [35]

Answer:

-25 and 2

Step-by-step explanation:

x + y = -23

xy = 50, so y = 50 /x

x + 50/x =-23, x^2 +23x - 50 = (x-2)(x+25)

so 2 and -25

4 0

3 years ago

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

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

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

$\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34$

So we end up with

$\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}$

###

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

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

4 0

3 years ago

Estimate the problem 3,844 ÷75

blondinia [14]

Answer:88

Step-by-step explanation:

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3 years ago

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Eleven less than three times a number

Zina [86]

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3 years ago

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Elena-2011 [213]

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3 years ago