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Gelneren [198K]
3 years ago
10

Morgan, Dakota and Taylor showed their work solving the equation 7x+5x+1=73. Identify the student with the invalid first step an

d justify your reasoning. Morgan: 7x+5x+1=73 12x+1=73 Dakota: 7x+5x+1=73 7x+6x=73 Taylor: 7x+5x+1=73 7x+5x=72
Mathematics
2 answers:
antoniya [11.8K]3 years ago
8 0

Answer:

Step-by-step explanation:

Given the equation solved by Morgan, Dakota and Taylor expressed as 7x+5x+1=73, to get the value of x, the valid steps are as shown;

7x+5x+1=73

take the sum of 7x and 5x

12x+ 1 = 73

subtract 1 from both sides

12x+1-1 = 73-1

12x = 72

Divide both sides by 12

12x/12 = 72/12

x = 6

Based on the calculation above, the students with invalid first step is Dakota.

<em>Dakota added 5x with 1 to get 6x which is an invalid step because 1 is not attached to a variable x. A function cannot be added to a constant hence making his first step invalid.</em>

mrs_skeptik [129]3 years ago
4 0

Answer:

As you can see, Taylor failed to add 1 to her equation. This ruins the whole first step. She also used 72 instead of 73 for the answer. Taylor is incorrect.

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Write an equation of the line with a slope of 2 that passes through the point (−1, 4) in point-slope form and slope-intercept fo
MrMuchimi

Answer:

y=2x+6

Step-by-step explanation:

y=mx+b

your two point that are [-1, 4] with 1 as x and 4 as y

replace y with 4 then replace m with 2 then put your negative one in () it should look like this

4=2(-1)+b then do what in Parentheses

that will give you -2 then add 2 to 4 and that will give u 6

y=2x+6

3 0
3 years ago
How do you factor 125-x^3?
xz_007 [3.2K]

Step-by-step explanation:

We have

125 - x {}^{3}

First, 125 is a perfect cube because

5 \times 5 \times 5 = 125

and

x^3 is a perfect cube because

x \times  x \times x = x {}^{3}

so we can use the difference of cubes identity

( {x}^{3}  -  {y}^{3} ) = (x - y)( {x}^{2}  + xy +  {y}^{2} )

Let say we have two perfect cubes:

64 because 8×8×8=64

and 27 because 3×3×3=27 and let subtract

64 - 27

we know that

64 - 27 = 37

but using the difference of cubes identity we should get the same thing.

Remeber cube root of 64 is 4 and cube root of 27 is 3 so we have

(4 - 3)( {4}^{2}  + 4(3) + 3 {}^{2} )

1(16 + 12 + 9) = 1(37) = 37

So the difference of cubes works for real numbers. This is a good way to help remeber the identity using real numbers.

Back on to the topic,

we know that 5 is cube root of 125 and x is the cube root of x^3 so we have

(5 - x)( {5}^{2}  + 5x +  {x}^{2} ) =

(5 - x)(25 + 5x +  {x}^{2} )

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3 years ago
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Step-by-step explanation:

4 0
3 years ago
If g (x) = 1/x then [g (x+h) - g (x)] /h
lys-0071 [83]

Answer:

\dfrac{-1}{x(x+h)}, h\ne 0

Step-by-step explanation:

If g(x) = \dfrac{1}{x}, then g(x+h) = \dfrac{1}{x+h}. It follows that

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

  = \dfrac{-1}{x(x+h)}, h\ne 0

5 0
3 years ago
I don’t know how to solve this
g100num [7]
The two equations graphs intersect and the points where they are touching are belonging to both graphs therefore solutions for both equations.

(2) points (x,y) are
(-1,0) (-1)^2. +0^2=1; 1=1 ✔️
0=-1+1; 0=0✔️
(0, 1). (0)^2. +1^2=1; 1=1 ✔️
1=0+1; 1=1 ✔️
6 0
3 years ago
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