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4 years ago

What's slope intercept form

Mathematics

2 answers:

djverab [1.8K]4 years ago

8 0

Slope intercept form: y=mx+b

dezoksy [38]4 years ago

6 0

Slope intercept form which is more commonly known as y = mx + b form is very helpful when graphing a line. The m or the coefficient of the x-term represents slope and the b or thee constant term represents the y-intercept.

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While standing in front of the school

kenny6666 [7]

Answer:

Step-by-step explanation:

what do u see!????????

4 0

3 years ago

Please Help Me With Matrices! Fully explain to me what to do, for I have no clue what I am doing. I can substitute, and eliminat

prohojiy [21]

Can you find an explanation of "row operations" with examples in any of your learning materials, online or in print?

Once you get the hang of row ops, it's not terribly hard. This does, however, take a lot of arithmetic.

−6x−y−5z=−10
− 5x+6y+4z=−7
2x−3y−2z=3

can be represented by the matrix

-6 -1 -5 -10
-5 6 4 -7
2 -3 -2 3

Our goal is to transform this 3 x 4 matrix so that it ends up looking like:

1 0 0 a
0 1 0 b
0 0 1 c

and the solution you want is the vector (a, b, c) (three numeric values).

I have more or less arbitrarily chosen to start with the third row:
2 -3 -2 3. We want this row to begin with a 1, so we multiply each of the original four digits by (1/2), obtaining 1 -3/2 -2/2 3/2, or 1 -3/2 -1 3/2.

We can present the original matrix in any order without changing its value. Thus, the original

-6 -1 -5 -10
-5 6 4 -7
2 -3 -2 3

becomes

-6 -1 -5 -10
-5 6 4 -7
1 -3/2 -1 3/2

We want that "1" to appear in the upper, left hand corner of the matrix. We are free to interchange rows, so we interchange the first and 3rd rows, obtaining

1 -3/2 -1 3/2
-5 6 4 -7
-6 -1 -5 -10

Next, we manipulate the first row (which begins with 1) so as to get the first element of the 2nd and 3rd rows to be 0.

To achieve this for the 2nd row, we multiply the 1st row by 5, obtaining

5 -15/2 -5 15/2

and then we add this to the existing 2nd row. The result will be an "0"
in the first column:

0 (6-15/2) ( 4-5) (-7+15/2), or 0 -3/2 -1 1/2.

Substitute this new 2nd row for the original 2nd row. We'll now have:

 1 -3/2 -1 3/2
0 -3/2 -1 1/2
-6 -1 -5 -10

Now we have to "fix" the 3rd row, so that it starts with a zero (0):
To accomplish this, mult. the first row by 6 and add the resulting new row to the existing 3rd row. Result should be 0 -10 -11 -1, and the revised matrix will be

1 -3/2 -1 3/2
 0 -3/2 -1 1/2
0 -10 -11 -1

Next steps involve transforming the 2nd column so that it looks lilke

0
1
0.

To do this, mult. the entire 2nd row by -2/3, Here's the expected result:

0 1 2/3 -1/3

Replace the existing 2nd row with this revised 2nd row:

1 -3/2 -1 3/2
 0 -3/2 -1 1/2
0 -10 -11 -1 becomes

1 -3/2 -1 3/2
0 1 2/3 -1/3
0 -10 -11 -1

In the end we want this matrix to look like

1 0 0 a
0 1 0 b
0 0 1 c

and the solution you want is the vector (a, b, c) (three numeric values).

Use this new 2nd row to further fix the 2nd column, so that it looks like

0
1
0.

I ask that you go thru this discussion and work out each set of calculations yourself, to verify what I have done so far. Reply with any questions that arise. We'll find a way to finish this solution.

8 0

3 years ago

Paula used 1 3/4 cups of flour for a recipe. How many cups y of flour did Paula have if she had x cups before making the recipe?

Svetach [21]

Y=x-1 3/4 all you have to do is subtract the amount used (1 3/4) form the starting amount (x)

4 0

3 years ago

Read 2 more answers

A rectangular piece of paper has a width that is 3 inches less than its length. It is cut in half along a diagonal to create two

Oliga [24]

Let

x-------> the length of the rectangle

y------> the width of the rectangle

we know that

The area of the rectangle is equal to

$A=x*y$

The area of the two congruent right triangles is equal to the area of the rectangle

$A=2*44=88\ in^{2}$

$88=x*y$ -------> equation A

$y=x-3$ -----> equation B

Substitute equation B in equation A

$x*[x-3]=88$

$x^{2} -3x-88=0$ --------> equation that can be used to solve for the length of the rectangle

Using a graph tool-------> solve the quadratic equation

see the attached figure

The solution is

$x=11\ in$ -----> the length of the rectangle

Find the value of y

$88=11*y$

$y=8\ in$ ----> the width of the rectangle

Statements

case A) The area of the rectangle is $88$ square inches

The statement is True

See the procedure

Case B) The equation $x*[x-3]=44$ can be used to solve for the dimensions of the triangle

The statement is False

Because, the equation $x*[x-3]=88$ can be used to solve for the dimensions of the triangle

case C) The equation $x^{2} -3x-88=0$ can be used to solve for the length of the rectangle

The statement is True

see the procedure

case D)The triangle has a base of $11$ inches and a height of $8$ inches

The statement is True

Because, the base of the triangle is equal to the length of the rectangle and the height of the triangle is equal to the width of the rectangle

case E) The rectangle has a width of $4$ inches

The statement is False

See the procedure

8 0

3 years ago

Read 2 more answers

Help. I’m stuck on this one skskksks

VMariaS [17]

I'm pretty sure the answer is: h = 9

Reason:

The question tells us that to complete the equation that represents the relationship between d and h, so for "d", its d = 1, for h, it is 9 per each "days"

from the hours.

6 0

4 years ago

Read 2 more answers