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2 years ago

9

Given the table below, determine which type of equation best models the data and use a calculator to find an equation of best fi

t.

1 answer:

irina1246 [14]2 years ago

4 0

Answer:

Hello,

Step-by-step explanation:

Best fit: quadratic y=5.3x²-9.3+6.1

with an total error of 0.4

You might be interested in

The measure on angle 0 is 7pie/4. Measure of its reference angle is __ °, and the tan 0 is ___

soldier1979 [14.2K]

Answer:

Reference angle is $\frac{\pi }{4}$ and tan $\frac{7\pi }{4}$ = -1

Step-by-step explanation:

$\frac{7\pi }{4}$ lies in the 4th quadrant

and reference angle for angle in 4th quadrant is 2π - x , where x is the angle given.

and reference angle for $\frac{7\pi }{4}$ is

2π- $\frac{7\pi }{4}$ = $\frac{\pi }{4}$

and tan $\frac{7\pi }{4}$ = -tan $\frac{\pi }{4}$ = -1

Since tan is negative in 4th quadrant ,therefore value will be negative for the given angle.

7 0

3 years ago

Write an equation of the line that is perpendicular to y = 1/2

atroni [7]

Answer:

the equation of the line that is perpendicular to y = 1/2x + 3 and passes through the point (10, -5)

= -5 = -2x + 15

Step-by-step explanation:

Write an equation of the line that is perpendicular to y = 1/2x + 3 and passes through the point (10, -5).

Using the slope intercept equation,

y = mx +c

m = slope = 1/2

For two lines to be perpendicular, the product of their slopes is -1

Let the slope of the other line be m2

m1×m2 =-1

1/2×m2 = -1

m2 = -1/(1/2) = -2

Slope of line = -2

For points (10, -5), x = 10, y =-5

-5 = -2× 10 +c

-5 = -20+ c

c = -5+20= 15

the equation of the line that is perpendicular to y = 1/2x + 3 and passes through the point (10, -5)

-5 = -2x + 15

6 0

3 years ago

Find c.<br> Round to the nearest tenth.

solmaris [256]

Answer: Picture is blury for me I can not see it well enough

Step-by-step explanation:

All i see is the 8 ft and 17ft cant see any of the other numbers

8 0

3 years ago

Please help!!!! ASAP giving brainiest

mariarad [96]

Answer:

$JK=2.25\ units$

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional, and this ratio is called the scale factor

so

$\frac{JK}{ST}=\frac{KL}{TU}$

substitute the given value and solve for JK

$\frac{JK}{1.5}=\frac{6}{4}$

$JK=(1.5)\frac{6}{4}$

$JK=2.25\ units$

8 0

2 years ago

Solve the given system of equations using either Gaussian or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTI

cricket20 [7]

Answer:

The system has infinitely many solutions

$\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}$

Step-by-step explanation:

Gauss–Jordan elimination is a method of solving a linear system of equations. This is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

An Augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

There are three elementary matrix row operations:

Switch any two rows
Multiply a row by a nonzero constant
Add one row to another

To solve the following system

$\begin{array}{ccccc}x_1&-3x_2&-2x_3&=&0\\-x_1&2x_2&x_3&=&0\\2x_1&+3x_2&+5x_3&=&0\end{array}$

Step 1: Transform the augmented matrix to the reduced row echelon form

$\left[ \begin{array}{cccc} 1 & -3 & -2 & 0 \\\\ -1 & 2 & 1 & 0 \\\\ 2 & 3 & 5 & 0 \end{array} \right]$

This matrix can be transformed by a sequence of elementary row operations

Row Operation 1: add 1 times the 1st row to the 2nd row

Row Operation 2: add -2 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1

Row Operation 4: add -9 times the 2nd row to the 3rd row

Row Operation 5: add 3 times the 2nd row to the 1st row

to the matrix

$\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]$

The reduced row echelon form of the augmented matrix is

$\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]$

which corresponds to the system

$\begin{array}{ccccc}x_1&&-x_3&=&0\\&x_2&+x_3&=&0\\&&0&=&0\end{array}$

The system has infinitely many solutions.

$\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}$

7 0

3 years ago