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2 years ago

What is the equation of this trend line? Enter your answers by filling in the boxes

Mathematics

1 answer:

Ymorist [56]2 years ago

4 0

Answer:

-1/2M + 110

Step-by-step explanation:

The slope of the line if the scale is right should be -1/2M as it goes down half the speed. The y intercept is 110.

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Ms. Kork sold her car for $8700. This was $100 more than two fifths of what she had paid for the car originally. How much mad Ms

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8600 is 100 less

3440?

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2 years ago

Why were there problems between the Spanish and apache.

Inessa05 [86]

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2 years ago

Point Z on a coordinate grid is at (-1,3). If the point is reflected across the line y=-x, what are the coordinates of its image

Neko [114]

(-3,1)

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3 years ago

There are 42 lorries, 54 cars and 284 motorbikes on a ferry.

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Step-by-step explanation:

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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

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

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

2 years ago