All categories

2 years ago

Write each equation in standard form using integers? y = 3x + 1 y = 4x - 7

Mathematics

1 answer:

Maru [420]2 years ago

3 0

Answer:

The standard form of y = 3 x + 1 is -3x + y = 1 .

The standard form of y = 4x - 7 is -4x + y = -7.

Step-by-step explanation:

Here the given equation are:

y = 3 x + 1

and y = 4x - 7

The above equations are of the form: y = m x + C

Now, the STANDARD FORM of the equation is Ax + By = C

<u>Consider equation 1: </u>

y = 3 x + 1

or, y - 3x = 1

or. -3x + y = 1

⇒ The equation -3x + y = 1 is in STANDARD FORM.

<u>Consider equation 2: </u>

y = 4 x - 7

or, y - 4x = -7

or. -4x + y = -7

⇒ The equation -4x + y = -7 is in STANDARD FORM.

You might be interested in

2. Check the boxes for the following sets that are closed under the given

son4ous [18]

The properties of the mathematical sequence allow us to find that the recurrence term is 1 and the operation for each sequence is

a) Subtraction

b) Addition

c) AdditionSum

d) in this case we have two possibilities

* If we move to the right the addition

* If we move to the left the subtraction

The sequence is a set of elements arranged one after another related by some mathematical relationship. The elements of the sequence are called terms.

The sequences shown can be defined by recurrence relations.

Let's analyze each sequence shown, the ellipsis indicates where the sequence advances.

a) ... -7, -6, -5, -4, -3

We can observe that each term has a difference of one unit; if we subtract 1 from the term to the right, we obtain the following term

-3 -1 = -4

-4 -1 = -5

-7 -1 = -8

Therefore the mathematical operation is the subtraction.

b) $0. \sqrt{1}. \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}$ ...

In this case we can see more clearly the sequence when writing in this way

$0, \sqrt{1^2}. \sqrt{2^2}, \sqrt{3^2 } . \sqrt{4^2} , \sqrt{5^2}$

each term is found by adding 1 to the current term,

$\sqrt{(0+1)^2} = \sqrt{1^2} \\\sqrt{(1+1)^2} = \sqrt{2^2}\\\sqrt{(2+1)^2} = \sqrt{3^2}\\\sqrt{(5+1)^2} = \sqrt{6^2}$

Therefore the mathematical operation is the addition

c) $... \frac{-10}{2}. \frac{-8}{2}, \frac{-6}{2}, \frac{-4}{2}. \frac{-2}{2}. ...$

The recurrence term is unity, with the fact that the sequence extends to the right and to the left the operation is

To move to the right add 1

- $\frac{-10}{2} + 1 = \frac{-10}{2} - \frac{2}{2} = \frac{-8}{2}\\\frac{-8}{2} + \frac{2}{2} = \frac{-6}{2}$

To move left subtract 1

$\frac{-2}{2} - 1 = \frac{-4}{2}\\\frac{-4}{2} - \frac{2}{2} = \frac{-6}{2}$

Using the properties the mathematical sequence we find that the recurrence term is 1 and the operation for each sequence is

a) Subtraction

b) Sum

c) Sum

d) This case we have two possibilities

If we move to the right the sum

If we move to the left we subtract

Learn more here: brainly.com/question/4626313

5 0

2 years ago

Solve the System of equations using substitution or elimination. Show work. 3x-4y=-31 X+y=6

Nataly_w [17]

Hope this helps I got (-1, 7)

6 0

3 years ago

The Spotlight on Teaching and Learning at the beginning of this chapter discusses the emphasis in the Common Core State Standard

ivanzaharov [21]

I don't know this question

3 0

3 years ago

consider the four corners of the front wall of a rectangular class room (upper right, upper left, lower left, upper left) which

professor190 [17]

(You copied 'upper left' twice, and you left out 'lower right'.
But we know what you mean.)

If those are the corners of the wall, then they're ALL in the plane of
the wall (coplanar with it).

3 0

3 years ago

What is missing from y+ 2= -3(x-4

Likurg_2 [28]

Answer:

The other parenthesis, but if that was a typo then I'm not sure. Hope this helps! :)

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers