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1 year ago

Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).

1 answer:

6 0

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5$

well then therefore

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}$

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}$

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T = days passed

r = rate of growth

by 0 day, or t = 0, there are 2 folks sick,

$\bf \qquad \textit{Amount for Exponential Growth}
\\\\
A=P(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &2\\
P=\textit{initial amount}\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &0\\
\end{cases}
\\\\\\
2=P(1+r)^0\implies 2=P\cdot 1\implies 2=P\qquad \boxed{A=2(1+r)^t}$

by the third day, t = 3, there are 40 folks sick,

$\bf \qquad \textit{Amount for Exponential Growth}
\\\\
A=P(1 + r)^t\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &40\\
P=\textit{initial amount}\to &2\\
r=rate\to r\%\to \frac{r}{100}\\
t=\textit{elapsed time}\to &3\\
\end{cases}
\\\\\\
40=2(1+r)^3\implies 20=(1+r)^3\implies \sqrt[3]{20}=1+r
\\\\\\
\sqrt[3]{20}-1=r\implies 1.7\approx r\qquad \boxed{A=2(2.7)^t}$

how many folks are there sick by t = 6? $\bf \stackrel{that~many}{A=2(2.7)^6}$

8 0

2 years ago

30 ounces at $4.00 per pound. how much was paid for 30 ounces

Nata [24]

It would be $7.50 because 4 ounces equals $1 so 30/4 =7.5

3 0

3 years ago

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

3 years ago

Determine the term life insurance amount per thousand on a 22-year-old female for a 5-year policy given that the face value of t

VLD [36.1K]

We need to determine the term life insurance amount per thousand on a 22-year-old female for a 5-year policy given that the face value of the policy is $600,000 and the annual premium is $1332.

Since we are supposed to find the term life insurance amount per thousand, therefore, we will use the following formula:

$\text{Term life insurance amount}=\frac{\text{Annual premium}}{\text{Face value of insurance}}*1000$

Upon substituting the values of annual premium and face value of insurance, we get:

$\text{Term life insurance amount}=\frac{\text{1332}}{\text{600000}}*1000\\
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Therefore, the required life insurance amount is $2.22.

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3 years ago

Read 2 more answers

Find the arc length, x, when

Arte-miy333 [17]

The arc length of the circle is 5π/9 units

<h3>How to determine the arc length?</h3>

From the question, we have the following parameters

Angle, ∅ = 5π/9

Radius, r = 1 unit

The arc length (x) is calculated as

x = r∅

Substitute the known values in the above equation

x = 5π/9 * 1

Evaluate the product

x = 5π/9

Hence, the arc length of the circle is 5π/9 units