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Triss [41]
3 years ago
9

Write an equation in point slope form for a line with a slope of -3 that passes through (2, -4)​

Mathematics
1 answer:
ruslelena [56]3 years ago
5 0

Answer:

work shown and pictured

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Before his basketball game, Jessie is responsible for cleaning the visiting teams team's half of the court. Three of his teammat
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It is D:1/6 because 3 of them do the home side and he cleans the other side.
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Scott has 18 3/4 gallons of gas in his tank. If he uses 3/50 gallons of gas per mile he drives how many miles can he drive befor
iVinArrow [24]

Answer:

Step-by-step explanation:

I’m sorry that i can’t fully help and that i answered it wrong twice i also got stuck. I hope someone else helps you and that you get an a

18 3/4 as an improper fraction is 75/4.

75/4 can be re-written as 1875/100 (multiply both sides by 25)

3/50 can be re-written as 6/100 (multiply both sides by 2)

6 0
3 years ago
Find the value of x in the figure below
joja [24]
I’m not sure but I added them up & combined like terms & got x=17.
6 0
2 years ago
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Henry is asked to find the exact value of cosine StartFraction 10 pi Over 3 EndFraction. His steps are shown below. 1. Subtract
lapo4ka [179]

Answer:

A. The reference angle should be pi/3, and the sign of the value should be negative.

Step-by-step explanation:

Just took the unit review.

3 0
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}

4 0
2 years ago
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