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Zinaida [17]
1 year ago
7

6x4=8xN how do i solve this equation

Mathematics
2 answers:
kow [346]1 year ago
5 0

Answer:

whats equivelent to 6x4?

Step-by-step explanation:

6 x 4 = 24

8 x 2 = 16 so 8 x 3 = 24 because 8 x 4 would be

8 x 2 = 16

16 x 2 =

6 + 6 = 12

carry the one.

2

1 + 1 + 1 (We now have three ones from carrying that 1) and our answer is

32.

So N = 3

12345 [234]1 year ago
5 0

Answer:

6x4=8xN N=8x3

Step-by-step explanation: N=8x3 because 6x4=24 so if you multiply 8x3 it =  

24

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Find a line that lies entirely in the set defined by the equation x2 + y2 − z2 = 1.
RUDIKE [14]

Answer:

x=1\,,\,y=t\,,\,z=t

Step-by-step explanation:

A line is a one-dimensional figure that has no thickness and extends infinitely in both directions.

A line segment is a line that has two end-points and a ray is a line that has one end-point.

Given equation is x^2+y^2-z^2=1

To find : a line that lies entirely in the set defined by the given equation.

Solution:

Take x=1\,,\,y=t\,,\,z=t

Check:

L.H.S=1^2+t^2-t^2=1+0=1=R.H.S

Therefore, x=1\,,\,y=t\,,\,z=t satisfy the given equation.

8 0
2 years ago
The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si
natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

                                g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

                                m = f'(x) = g'(x_o)

- We will develop the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o   ,     x_o = 10x + 2    

- For x_o = 10x + 2  ,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Solve the quadratic equation above:

                                 x = -0.0574, -0.387      

- Largest slope is at x = -0.387 where equation of line is:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- Other tangent line:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

6 0
1 year ago
Which of the following are terms in the sum of the given polynomials? 3x2,-2x2y, 2xy2, 4y2, xy2, -y2,2x2y
vampirchik [111]
Create the sum of the given polynomial terms.
f(x,y) = 3x² - 2x²y + 2xy² + 4y² + xy² - y² + 2x²y
        = 3x² + 3xy² + 3y²

Test the given terms to see if they belong in f(x,y).
3:       NO
3x²:   YES
3y²:   YES
3xy²: YES
4x²:   NO

Answer:
The terms in the sum of the given polynomials are 3x², 3y², and 3xy².

6 0
2 years ago
Selma’s class is making care packages to give to victims of a natural disaster. Selma packs one box in 5 minutes and has already
luda_lava [24]
I'm not sure what your asking, but if your asking how long it took her, it took her one hour. There is 12 boxes, and each one took her 5 minutes. 12*5 is equal to 60.
6 0
2 years ago
What are the vertical asymptotes of the function f(x) = the quantity of 4x plus 8, all over x squared plus 3x minus 4? plz answe
kolezko [41]

Answer:

x=1 and x=-4

Step-by-step explanation:

Put it into your calculator, go to graph and look at which points it says ERROR.

4 0
2 years ago
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