All categories

4 years ago

Determine if the equation is parallel, perpendicular, or neither;

2 answers:

6 0

The equations are parallel because the slopes are opposite reciprocals, which means if you flip -3/2 and multiply by -1 you get 2/3.

sweet [91]4 years ago

6 0

To tell is an equation is parallel or perpendicular, you don't have to do any math! Just look at the numbers! A parallel line will have the same slope or number before X. So these lines don't have the same slope so it's not parallel. You can tell if a line is perpendicular if the slopes are reversed. -3/2 is the opposite of 2/3 so it's perpendicular. Hope I helped!

You might be interested in

write an equation in point slope form of the line that passes through the point (5/6 , 0) and has a slope of m=-3

lyudmila [28]

$(\stackrel{x_1}{\frac{5}{6}}~,~\stackrel{y_1}{0})~\hspace{10em} \stackrel{slope}{m}\implies -3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-3}(x-\stackrel{x_1}{\frac{5}{6}})$

8 0

2 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

Find the solution 4x<8

Nataliya [291]

x=1

...................

7 0

3 years ago

What is the equation of the line in standard form?<br> 3x−y=−6 3x + y = 6 x + 6y = 9 x−6y=−9

lisabon 2012 [21]

The standard form: Ax + By = C

3x - y = -6 YES - A = 3, B = -1, C = -6

3x + y = 6 YES - A = 3, B = 1, C = 6

x + 6y = 9 YES - A = 1, B = 6, C = 9

x - 6y = -9 YES - A = 1, B = -6, C = -9

8 0

3 years ago

What do you need to know?

cupoosta [38]

Answer:

incorrect

Step-by-step explanation:

the volume of a cube = s³ ( where s is the length of the side )

to find s take the cube root

s = $\sqrt[3]{27}$ = 3

since 3 × 3 × 3 = 27

4 0

4 years ago