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3 years ago

8

I’m so lost how to answer what x and y equals to??

1 answer:

-Dominant- [34]3 years ago

4 0

Short answer: X=46°
Y=134°

So the line represents 180° since it is exactly half of 360°, the line is separated by two angles, an obtuse angle and an acute angle, we already know the value of the obtuse angle which is 134°, to find the degree of the acute angle we do:
180-134
180-134=46
The acute angle on the opposite side is 46°. Now the line intersecting the angles acts as a guide and if you look closely you can see that the angle on the second line is an inverted version of the first line. Seeing that you can now infer that Y=134° and X on the bottom line equals 46°. I hope this wasn’t too confusing since I tend to over explain myself.

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3 years ago

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What does x equal?. 3/4 x + 5/8=4x

Aleonysh [2.5K]

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3 years ago

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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

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

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

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

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2 years ago

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hammer [34]

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3 years ago

18) through: (-3,-3), parallel to y = -x + 3

otez555 [7]

Step-by-step explanation:

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3 years ago