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KonstantinChe [14]

2 years ago

13

Which equation represents a line that passes through (–9, –3) and has a slope of –6?

1 answer:

brilliants [131]2 years ago

8 0

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ -9}}\quad ,&{{ -3}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies -6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-3)=-6[x-(-9)]
\\\\\\
y+3=-6(x+9)\implies y+3=-6x-54\implies y=-6x-57$

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Nikolay [14]

Answer:

-88

Step-by-step explanation:

Let N(x) be the number of students at NMS for a certain number of years.

Let S(x) be the number of students at SMS for a certain number of years.

Let x be the number of years.

N(x)=1240-25x

S(x)=800-30x

When # of students is equal, N(x)=S(x). Therefore, we are looking for the value of x (# of years) when 1240-25x=800-30x

Subtract 1240 from each side

-25x=-440-30x

Add 30x

5x=-400

x=-88 years

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Answer:

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Step-by-step explanation:

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If log2 5 = k, determine an expression for log32 5 in terms of k.

lukranit [14]

Answer:

$log_3_2(5)=\frac{1}{5} k$

Step-by-step explanation:

Let's start by using change of base property:

$log_b(x)=\frac{log_a(x)}{log_a(b)}$

So, for $log_2(5)$

$log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)$

Now, using change of base for $log_3_2(5)$

$log_3_2(5)=\frac{log(5)}{log(32)}$

You can express $32$ as:

$2^5$

Using reduction of power property:

$log_z(x^y)=ylog_z(x)$

$log(32)=log(2^5)=5log(2)$

Therefore:

$log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)$

As you can see the only difference between (1) and (2) is the coefficient $\frac{1}{5}$ :

So:

$\frac{log(5)}{log(2)} =k\\$

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2 years ago

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lesya [120]

Answer:

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Step-by-step explanation:

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