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4 years ago

4x^2 – 19x – 5 = 0

Mathematics

1 answer:

Ann [662]4 years ago

8 0

Answer:

Step-by-step explanation:

4x² - 19x -5 =0

multiply coefficient of x² and the constant.

here coefficient of x² = 4 and constant = -5

4*-5 = -20

Coefficient of x = -19

so, when you multiply you have to get (-20) and when you subtract you should get -19

(-20)*1= -20

-20 + 1 = -19

4x² - 19x - 5 = 0

4x² - 20x + x - 1*5 = 0

4x(x - 5) + (x - 5) = 0

(x - 5)(4x + 1) = 0

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2x – yl. when x = 3 and y = 10

iragen [17]

Answer:

Step-by-step explanation:

Here you go mate

Step 1

2x-y Equation/Question

Step 2

2x-y Evaluate for x and y

2(3)-10

Step 3

2(3)-10 Simplify

6-10

answer

-4

Hope this helps

5 0

3 years ago

A bar graph shows that sports books received 9 votes. If that scale is 0 to 20 by twos, where should the bar end for the sports

vladimir1956 [14]

I'm pretty sure it would go between 8 and 10 if it's 9 sports books it would go between 8 an 10

8 0

3 years ago

What is an equation for the line perpendicular to y = 3x – 1 that passes through (-9, -2)?

FinnZ [79.3K]

Answer:

y = -1/3x - 5

Step-by-step explanation:

Perpendicular lines have negative reciprocal slopes, so the line is:

y = -1/3x + b

Finding the value of b, using the given point (-9, -2)

-2 = -1/3*(-9) + b
-2 = 3 + b
b = -2 - 3
b = -5

The line is

y = -1/3x - 5

3 0

3 years ago

In Professor Friedman's economics course the correlation betweenthe students' total scores before the final examination and thei

GalinKa [24]

Answer:

slope=0.16

intercept=30.2

78.2

36%

Step-by-step explanation:

Question 1

We are given that

xbar=280

sx=30

ybar=75

sy=8

r=0.6

The regression line can be written as

y=a+bx

a=intercept

b=slope

where

$b=r\frac{sy}{sx}$

and

$a=ybar-bxbar$

b=0.6*(8/30)

b=0.16

a=75-0.16*280

a=30.2

Thus,

slope=0.16

intercept=30.2

Question 2

The regression line in the given scenario

y=30.2+0.16x

Julie pre exam total before the exam was 300.

y=30.2+0.16*300

y=30.2+48=78.2

So, the predicted final exam score of Julie is 78.2.

Question 3

R² denotes the variation in dependent variable y explained by the linear relationship of x and y.

R²=0.6²=0.36

Thus, the proportion of the variation in final exam scores that is explained by the linear relationship between pre-exam scores and final exam scores is 36%.

5 0

4 years ago

Determine whether the given system has a unique solution, no solution, or infinitely many solutions.

victus00 [196]

Answer:

The system has no solution.

Step-by-step explanation:

To find the solution to this system of linear equations

$\left\begin{array}{ccccc}-3x_1&+x_2&-2x_3&=&8&\\x_1&+5x_2&-x_3&=&4&\\-x_1&+11x_2&-4x_3&=&1&\end{array}\right$

First, state the problem in matrix form, this means, extracting only the numbers, and putting them in a box.

$\left[ \begin{array}{ccc|c} -3 & 1 & -2 & 8 \\\\ 1 & 5 & -3 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

This is called an augmented matrix. The word “augmented” refers to the vertical line, which we draw to remind ourselves where the equals sign belong

Next, transform the augmented matrix to the reduced row echelon form with the help of Row Operations.

Row Operation 1: multiply the 1st row by -1/3

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 1 & 5 & -1 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

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

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

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

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ 0 & \frac{32}{3} & - \frac{10}{3} & - \frac{5}{3} \end{array} \right]$

Row Operation 4: multiply the 2nd row by 3/16

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]$

Row Operation 5: add -32/3 times the 2nd row to the 3rd row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]$

Row Operation 6: multiply the 3rd row by -1/15

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 7: add -5/4 times the 3rd row to the 2nd row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 8: add 8/3 times the 3rd row to the 1st row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 9: add 1/3 times the 2nd row to the 1st row

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

The reduced row echelon form of the augmented matrix is

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

which corresponds to the system

$\left\begin{array}{ccccc}x_1&&+\frac{9}{16} x_3&=&0&\\&1x_2&-\frac{5}{16}x_3&=&0&\\&&0&=&1&\end{array}\right$

Equation 3 cannot be solved, therefore, the system has no solution.

7 0

4 years ago