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3 years ago

X = 2 - y3x - 10 = yi have to find yplease help

Mathematics

1 answer:

drek231 [11]3 years ago

3 0

I think the first one is 2y and the secound is,

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3^2+4^2-(4x2) asdfsdfs

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

3^2+4^2-(4x2)

9+16-8

=17

4 0

3 years ago

Graph y<4x-3 idk which one it is

pav-90 [236]

Answer:

The bottom right graph is correct, with the y-intercept at -3 and the graph shaded to the right.

Step-by-step explanation:

The inequality of y < 4x - 3 is written in slope-intercept form (y = mx + b), where m = slope and b = y-intercept. The difference between a linear equation and linear inequality is that an equation only produces one answer while an inequality and can have several answers. Because the y-intercept = -3, the graph should cross the y-axis at -3, which it does for all the given graphs. However, the slope is 4, which means that the change in 'y' over the change in 'x' is 4/1 - up four units and to the right one unit. This would give us the choice of the bottom to graphs.

Since the inequality is y < 4x - 3, then all answers must be less than the total value of the expression, which means we shade to the right of the line.

4 0

4 years ago

Todd wants to find 352 + 116. Break

jekas [21]

Answer

given,

352 + 116

we have to break 116 and use it on open number line

breaking of number 116

116 = 100 + 10 + 6

And addition will be shown in the number line attached below

now,

352 + 100 = 452

452 + 10 = 462

462 + 6 = 468

we know

sum of

352 + 116 = 468

8 0

3 years ago

The following 3 points are on a parabola defining the edge of a ski.

Andrej [43]

a. The linear equation for the first point (-4,1) is 16A-4B+C=1

b. The linear equation for the second point (-2, 0.94) is 4A-2B+C=0.94

c. The linear equation for the third point (0,1) is 0A+0B+C=1

d. The matrix equation looks like this:

$\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right]$

e. The inverse of the coefficient matrix looks like this:

$A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]$

f. The equation of the parabola is: $\frac{3}{200}x^{2}+\frac{3}{50}x+1=y$

a. In order to build a linear equation from the given points, we need to substitute them into the general form of the equation.

Let's take the first point (-4,1). When substituting it into the general form of the quadratic equation we end up with:

$(-4)^{2}A+(-4)B+C=1$

which yields:

$16A-4B+C=1$

b. Let's take the second point (-2,0.94). When substituting it into the general form of the quadratic equation we end up with:

$(-2)^{2}A+(-2)B+C=0.94$

which yields:

$4A-2B+C=0.94$

c. Let's take the third point (0,1). When substituting it into the general form of the quadratic equation we end up with:

$(0)^{2}A+(0)B+C=1$

which yields:

$0A+0B+C=1$

d. A matrix equation consists on three matrices. The first matrix contains the coefficients (this is the numbers on the left side of the linear equations). Make sure to write them in the right order, this is, the numbers next to the A's should go on the first column, the numbers next to the B's should go on the second column and the numbers next to the C's should go on the third column.

The equations are the following:

16A-4B+C=1

4A-2B+C=0.94

0A+0B+C=1

So the coefficient matrix looks like this:

$\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]$

Next we have the matrix that has the variables, in this case our variables are the letters A, B and C. So the matrix looks like this:

$\left[\begin{array}{c}A\\B\\C\end{array}\right]$

and finally the matrix with the answers to the equations, in this case 1, 0.94 and 1:

$\left[\begin{array}{c}1\\0.94\\1\end{array}\right]$

so if we put it all together we end up with the following matrix equation:

$\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right]$

e. When inputing the coefficient matrix in our graphing calculator we end up with the following inverse matrix:

$A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]$

Inputing matrices and calculating their inverses depends on the model of a calculator you are using. You can refer to the user's manual on how to do that.

f. Our matrix equation has the following general form:

AX=B

where:

A=Coefficient matrix

X=Variables matrix

B= Answers matrix

In order to solve this type of equations, we can make use of the inverse of the coefficient matrix to end up with an equation that looks like this:

$X=A^{-1}B$

Be careful with the order in which you are doing the multiplication, if A and B change places, then the multiplication will not work and you will not get the answer you need. So when solving this equation we get:

$\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]*\left[\begin{array}{c}1\\\frac{47}{50}\\1\end{array}\right]$

(Notice that I changed 0.94 for the fraction 47/50 you can get this number by dividing 94/100 and simplifying the fraction)

So, in order to do the multiplication, we need to multiply each row of the coefficient matrix by the answer matrix and add the results. Like this:

$\frac{1}{8}*1+(-\frac{1}{4})(\frac{47}{50})+\frac{1}{8}*1$