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Hello Guest!

What is 2x-6?

A suggestion:

2x - 6 = 2 * (x - 3)

The term as a function
and the zero-point:

f(x) = 2x - 6 = 0

2x = 6

x = 3

Greeting asinus :- )

  !

[y² x⁶ ] /  [y³ x⁶ ]   =   1 / y

A line passes through the point  (-3, -1) with a slope of -3/2.

What is the equation of the line perpendicular to that line that passes through that point?

Lines that are perpendicular have slopes that are negative reciprocals.

So, if the first line has a slope of  -3/2  the line perpendicular to it will have a slope of  2/3.

Using the point slope form of a line for the line that passes through the point (-3, -1) with a slope of  2/3:

y - y₁  =  m(x - x₁)     --->     y - -1  =  (2/3)(x - -3)     --->     y + 1  =  (2/3)(x + 3)

You can simplify that equation:     y + 1  =  (2/3)x + (2/3)(3)

                                                     y + 1  =  (2/3)x + 2

                                                           y  =  (2/3)x + 1

The usual formula for the area of a cirlce is:  A  =  pi·r²

Since the clock face has a diameter of 20 centimeters, it will have a radius of 10 centimeters.

--->     The area of the clock face will be:  A  =  pi·10²     --->     A  =  100pi

Since there are 12 cngruent sectors on the clock face, the area of each sector is:  100pi / 12  =  25pi / 3

ctr+shift+reload=rotate 90°

ctr+alt+shift+reload=barrel role

or go to settings to change angle

search do a barrel role google to make google do a barrel role

Solve for z:  x  =  y / z

Multiply both sides by z:  x · z  =  y

Divide both sides by x:         z  =  y / x

Two points on the graph are  (-2, -2)  and (2, -5)

And the slope is given by :

[ - 5 - (-2)] / [ 2 - (-2) ]  =    -3 / 4

lol true

so please help

Solve:  log₄(x - 3) + log₄(x + 3)  =  2

--->          log₄ [ (x - 3)(x + 3) ]  =  2

--->                      (x - 3)(x + 3)  =  4²

--->                                 x² - 9  =  16

--->                                      x²  =  25

--->                                       x  =  5          (The possible solution -5 doesn't check.)

idk any of those lol

Solve for b:  h  =  -b / (2a)

Multiply both sides by 2a:     h·2a  =  -b

                                 --->       2ah  =  -b

Multiply both sides by -1:      -2ah  =  b

Factor:  5x² - 25x

Begin by factoring out the common term 5x

     5x² - 25x  =  5x(x - 5)

This is all that can be done in this problem.

Photomath

OK, EinsteinJr. Will take a kick at it!.

Since 1 mole of any element has=6.0221415e+23 atoms, then 3 moles will contain:

6.0221415e+23 x 3 =1.80664245e+24 atoms

1) After one half-life, it should have left=[1.80664245e+24] / 2 =9.03321225e+23 atoms.

2) After 2 half-lives, it should have left =[1.80664245e+24] x 2^-2=4.516606125e+23 atoms.

3)After 3.5 half-lives,,,,,,,,,,,,,,,,,,,,,,,,,,,,,=[1.80664245e+24] x 2^-3.5=1.596861e+23 atoms.

4) After 1 billion years,,,,,,,,,,,,,,,,,,,,,,,,,,=[1.80664245e+24] x 2^-4,073.32=0 atoms.

5) After 2.45 x 10^8 years,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,=0 atoms.

P.S. With the number of atoms that we start with(1.81 x 10^24) and a half-life of only 245,500 years, then it takes less than 80 half-lives or just under 20 million years for the complete decay of the element.

m = d-n / 3

F(t)= 6.3e^(0.027t), If the exponent of your function is as it is written here, then:

1) 6.3 mm^3 initial size of the fungus.

2) 10.811 mm^3 size after 20 days

3) 102.393 days required for the fungus to grow to 100 mm^3.

F(t)= 6.3e^(0.027)*t, If the exponent of your function is as it is written here, then:

1) 6.3 mm^3 initial size of the fungus.

2) 129.448 mm^3 size after 20 days.

3) 15.450 days needed for fungus to grow to 100 mm^3

My favorite thing about school is that it's almost over.

Mrs. Smith will have accumulated a total of $17,721.02 @ 2.8% comp. continuously for 25 years. Contrast this with the same investment by Mr. Appleton and you will see there is very little difference between their two investments.

5/8 + 1/3 =15/24 + 8/24=23/24 of a mile.

1.6 per min

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