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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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solong [7]
Well, 18 yards is 54 feet.
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3 years ago
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Darlene has 59 pages to type. so far she gas type 23 pages. what fraction of the job is finished ​
coldgirl [10]

Answer:

\frac{23}{59}

Step-by-step explanation:

A fraction can be found be taking the current amount over the total amount:

\frac{current}{total}=\frac{23}{59}

Since Darlene has only typed 23 pages out of 59, her fraction is 23 over 59.  Fractions must always be in simplest form, which means that both the numerator and denominator can no longer be divided by the same factor.  Since the numbers 23 and 59 are both prime and have no other factors other than 1 and themselves, the fraction 23/59 is in simplest form.

4 0
3 years ago
A transformation T : (x, y) → (x + 3, y + 1). Find the preimage of the point (4, 3) under the given transformation. (7, 4) (1, 2
motikmotik

Answer:

(1, 2)

Step-by-step explanation:

Remember that the final shape and position of a figure after a transformation is called the image, and the original shape and position of the figure is the pre-image.

In our case, our figure is just a point. We know that after the transformation T : (x, y) → (x + 3, y + 1), our image has coordinates (4, 3).

The transformation rule T : (x, y) → (x + 3, y + 1) means that we add 3 to the x-coordinate and add 1 to the y-coordinate of our pre-image. Now to find the pre-image of our point, we just need to reverse those operations; in other words, we will subtract 3 from the x-coordinate and subtract 1 from the y-coordinate.

So, our rule to find the pre-image of the point (4, 3) is:

T : (x, y) → (x - 3, y - 1)

We know that the x-coordinate of our image is 4 and its y-coordinate is 3.

Replacing values:

                (4 - 3, 3 - 1)

                (1, 2)

We can conclude that our pre-image is the point (1, 2).

6 0
2 years ago
.
Pachacha [2.7K]

Answer:

this is too old

Step-by-step explanation:

7 0
3 years ago
​Find all roots: x^3 + 7x^2 + 12x = 0 <br> Show all work and check your answer.
Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

<u>Solution:</u>

We have been given a cubic polynomial.

x^{3}+7 x^{2}+12 x=0

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

x^{3}+7 x^{2}+12 x=0

x\left(x^{2}+7 x+12\right)=0   ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

x^{2}+7 x+12=0

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}

<em><u>Therefore, the two roots are:</u></em>

\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}

And,

\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0
3 years ago
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