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3 years ago

Need help you guys 2.13X10^-5

Mathematics

1 answer:

tresset_1 [31]3 years ago

8 0

0.0000213, that should be it.

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Chole spent a night at her friend's house. She left at 7.00pm . She soon came back home 12 hours later . How many total of hours

DIA [1.3K]

Ok so we have 7:00 pm and she spends 12 more hours over.
Now lets count up 12 hours
=19
You could also just do 7+12! :)
So therefore, she spend 19 hours had her friends house! :D
Hope this helps!

6 0

3 years ago

Which of the following circles have their centers on the x-axis? Check all that apply.

Alexeev081 [22]

Answer: Option A and Option D

Step-by-step explanation:

By definition the general equation of a circle has the following form

$(x-h)^2+ (y-k)^2=r^2$

Where r is the radius and the point (h, k) is the center of the circle

All points that are on the x axis have the following form:

(h, 0)

Therefore all the circles that have their centers on the x axis will have a value of k = 0.

Identify among the options all those equations that have a value of k = 0

The circles that have their center on the x axis are option A and option D

7 0

3 years ago

the lengths of 3 pieces of wire are in the ratio of wire are in the ratio 10:15:8 if the length of the shortest piece of wire is

galina1969 [7]

3:4,5:2.4 2.4/4*5= 3 (8/4= 2 2*5=10) and 3/2*3= 4,5 (10/2*3=15)

5 0

3 years ago

Find the equation for the line that passes through the point (5,-3), and that is perpendicular to the line with the equation y-1

8090 [49]

Answer:

y+3=-4(x-5)

Explanation:

Part A

Given the line:

$y-1=\frac{1}{4}(x-2)$

We want to find the equation of a perpendicular line that passes through the point (5,-3),

First, determine the slope of the perpendicular line.

Comparing the given line with the slope-point form:

$\begin{gathered} y-y_1=m(x-x_1) \\ \implies\text{Slope},m=\frac{1}{4} \end{gathered}$

By definition, two lines are perpendicular if the product of their slopes is -1.

Let the slope of the perpendicular line = n

$\begin{gathered} \implies\frac{1}{4}n=-1 \\ n=-4 \end{gathered}$

Thus, using a slope of -4 and a point (5,-3), we find the equation of the line.

$\begin{gathered} y-y_1=m(x-x_1) \\ y-(-3)=-4(x-5) \\ y+3=-4(x-5) \end{gathered}$

The equation of the perpendicular line in the slope-point form is:

$y+3=-4(x-5)$

Part B

In order to graph the line, first, find two points on the line.

When x=0

$\begin{gathered} y+3=-4(0-5) \\ y+3=20 \\ y=20-3=17 \\ \implies(0,17) \end{gathered}$

When y=1

$\begin{gathered} 1+3=-4(x-5) \\ 4=-4(x-5) \\ \frac{4}{-4}=x-5 \\ -1=x-5 \\ x=5-1=4 \\ \implies(4,1) \end{gathered}$

Join the points (0,17) and (4,1) as shown in the graph below:

3 0

1 year ago

Please please please help me

Sunny_sXe [5.5K]

B is the answeri!!!!!

3 0

3 years ago