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igor_vitrenko [27]
3 years ago
15

What is a negative slope and passes through the origin?

Mathematics
1 answer:
KIM [24]3 years ago
7 0

Answer:

A negative slope is for the line running in opposite ways (downwards). And it cannot pass through the origin.

{ \underline{ \mathfrak{ \green{Good \: grace}}   \:  \:⚜ \:  \:  { \blue{ \mathfrak{Good \: God}}}}}

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Find the inverse function of
Neko [114]

The correct answer is C. m-1(x) = \frac{x + 5}{5}


You can find the value of any inverse function by switching the m(x) and the x value. Then you can solve for the new m(x) value. The end result will be your new inverse function. The step-by-step process is below.


m(x) = 5x - 5 ----> Switch m(x) and x

x = 5m(x) - 5 ----> Add 5 to both sides

x + 5 = 5m(x) ----> Divide both sides by 5

\frac{x + 5}{5} = m(x) ----> Change the order for formatting purposes.

m(x) = \frac{x + 5}{5}


And this would be your inverse, which matches answer C.



5 0
3 years ago
Which of the following properly describe “slope”?
oksian1 [2.3K]

Answer:

#3    1/2

Step-by-step explanation:

if you start where the line and the boxes in the graph line up like at (0,-2) and then you go up 1 and right 2 the line will be there again line up with the boxes and since slope is rise/run it is 1/2

7 0
4 years ago
A rectangle has a length of (2c + 1) and a width of (3 + c). If c = 3, what is the area of the rectangle?
Varvara68 [4.7K]

Answer:

Area of a rectangle = 42 square units

Step-by-step explanation:

Area of a rectangle = length times width.

l = (2c + 1)  when c = 3, the length = 2 · 3 + 1 ; l = 6 + 1 = 7

w = (3 + c) when c = 3, the width = 3 + 3; w = 6

Area of a rectangle = length times width

Area of a rectangle = 7 · 6

Area of a rectangle = 42 square units

7 0
2 years ago
Lify 8<br> 7x2-x+8 – 5x2 + 3x – 10
zalisa [80]

Answer: (-5x2 - 8x + 7) - (5x2 - 4)

Step-by-step explanation:

= -5x2 - 8x + 7 - 5x2 + 4

= -5x2 - 5x2 - 8x + 7 + 4

= -10x2 – 8x + 11

Answer: = (-5x2 - 8x + 7) - (5x2 - 4)

6 0
3 years ago
If alpha and beta are the zeroes of quadratic polynomial ax^2+bx+c , then find alpha^4+beta^4
weqwewe [10]
Ill write A for alpha and B for beta.

AB = c/a  and A + B = -b/a

A^4 + B^4 = (A^2 + B^2)^2  -   2A^2B^2

                 = [(A + B)^2 - 2AB] ^2 - 2A^2B^2

Plugging in the values for A+B and AB we get

A^4 + B^4 = [(-b/a)^2 - 2c/a]^2 - 2(c/a)^2

   =    (b^2 / a^2 - 2c / a)^2 - 2c^2/a^2

  = (b^2 - 2ac)^2  - 2c^2
     ----------------      -----
          a^4                a^2

 =   (b^2 - 2ac)^2 - 2a^2c^2
       -----------------------------
                   a^4
            
5 0
3 years ago
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