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1 year ago

Consider the equations y = 3x and y = -12- x.

Mathematics

1 answer:

jasenka [17]1 year ago

5 0

Answer:

x | -7 -5 -3

y | -21 -15 -9

Step-by-step explanation:

y = 3x

If x = -7 then y = 3 × (-7) = -21

x = -5 then y = 3 × (-5) = -15

x = -3 then y = 3 × (-3) = -9

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Paul and Tom both commute to work. Paul's commute on the train takes 15 minutes more than one third as many minutes as Tom's com

Ronch [10]

Answer:

45 = (one third)x + 15

Step-by-step explanation:

If x represents the number of minutes Tom commutes, then (1/3)x is "one-third as many minutes as Tom's commute." 15 minutes more than that is ...

(1/3)x + 15

We are told this is the length of Paul's commute, and that it is 45 minutes. So, the appropriate equation is ...

45 = (1/3)x +15

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2 years ago

Help 6th grade math please help i will give brainliest

lilavasa [31]

Answer:

multiply 12 times 9

Step-by-step explanation:

I'm pretty sure this is right, I'm sorry if it's wrong

3 0

2 years ago

Can someone help me in #66

VladimirAG [237]

$\bf log_{{ a}}(xy)\implies log_{{ a}}(x)+log_{{ a}}(y)
\\ \quad \\
% Logarithm of rationals
log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)
\\ \quad \\
% Logarithm of exponentials
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-------------------------------\\\\
$

$\bf log_4\left( \cfrac{\sqrt{x^5y^7}}{zw^4} \right)\implies log_4(\sqrt{x^5y^7})-log_4(zw^4)
\\\\\\
log_4\left[(x^5y^7)^{^\frac{1}{2}}\right]-log_4(zw^4)\implies 
\cfrac{1}{2}log_4\left[(x^5y^7)\right]-log_4(zw^4)
\\\\\\
\cfrac{1}{2}\left[ log_4(x^5)+log_4(y^7) \right] -[log_4(z)+log_4(w^4)]
\\\\\\
\cfrac{1}{2}log_4(x^5)+\cfrac{1}{2}log_4(y^7)-[log_4(z)+log_4(w^4)]$

$\bf \cfrac{1}{2}\cdot 5log_4(x)+\cfrac{1}{2}\cdot 7log_4(y)-log_4(z)-4log_4(w)
\\\\\\
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2 years ago

Use the figure below to determine the length of a to the nearest tenth.

Mademuasel [1]

Ion know about this I don’t know the answer fasho

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2 years ago

Can someone please answer the following questions step by step?

Juliette [100K]

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2 years ago