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krok68 [10]
2 years ago
6

I need help, pls and thx

Mathematics
1 answer:
Sauron [17]2 years ago
7 0

The solutions for the quadratic equation are given as follows:

x = -1, x = 7/5

<h3>What is a quadratic function?</h3>

A quadratic function is given according to the following rule:

y = ax^2 + bx + c

The solutions are:

  • x_1 = \frac{-b + \sqrt{\Delta}}{2a}
  • x_2 = \frac{-b - \sqrt{\Delta}}{2a}

In which:

\Delta = b^2 - 4ac

For this problem, the equation is:

5x² - 2x - 7 = 0.

Hence the coefficients are a = 5, b = -2 and c = -7, and then the solutions are found as follows:

  • \Delta = (-2)^2 - 4(5)(7) = 144
  • x_1 = \frac{2 + \sqrt{144}}{10} = \frac{7}{5}
  • x_2 = \frac{2 - \sqrt{144}}{10} = -1

The solutions are:

x = -1, x = 7/5

More can be learned about quadratic equations at brainly.com/question/24737967

#SPJ1

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mariarad [96]

Answer:

About 1 gal  

Step-by-step explanation:

When you are estimating with fractions, you should round them off to the closest integers that you can work with in your head.

⅓, ⅕, and ½ are all close to zero, but that's absurd, because Neil clearly has some paint .

Instead, let's estimate that ⅓ gal + ⅕ gal  make about ½ gal.

They still have the other ½ gal, and  

½ gal + ½ gal = 1 gal.

Neil has about 1 gal of paint.

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Step-by-step explanation:

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3 years ago
a line passes through the points (14,5) and (19,15) determine and equation for a perpendicular line that passes through the poin
PSYCHO15rus [73]
The equation of the line thru <span> (14,5) and (19,15)  is found as follows:

                      15-5           5
slope = m = ----------- = ------ = -1
                       19-14         5

Thus, the slope of any line perpendicular to this first line has the slope

m = -1/1 = -1.

Then, using the point-slope form, y-15 = -1(x-19), or y -15 = -x + 19, or
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4 years ago
Can someone help me with this question?
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4 years ago
4. Parking fees at IIUM are RM 5.00 for IIUM students and RM 7.50 for non-IIUM students. At the
professor190 [17]

Answer:

(a) 780 students and 960 non-students  

(b) No. The maximum revenue is RM9000 from 1200 non-students.

(c). Revenue is maximum of RM9000 at 1200 non-students, decreasing by RM2.50 per student to a minimum of RM6000 at 1200 students

Step-by-step explanation:

 Let x = IIUM students and

and  y = non-IIUM students

You have two conditions

(a)          x +         y = total vehicles parked

(b) 5.00x + 7.50y = total gross receipts

(a) Wednesday

From your table,  

       x +          y = 1740

5.00x + 7.70y = RM11 100  

Solve the simultaneous equations

\begin{array}{rrcrl}(1) & x + y & = &1740&\\(2) & 5.00x + 7.50y & = & 11 100\\(3)&  5.00x + 5.00y & = & 8700 & \text{Multiplied (1) by 5}\\&2.50 y & = &2400 &\text{Subtracted (3) from (2)}\\(4)&y& = &\mathbf{960} &\text{Divided each side by 2.50}\\& x +960& = &1740& \text{Substituted (4) into (1)}\\& x& = &\mathbf{780}& \\\end{array}\\\text{There are $\large \boxed{\textbf{780 students and 960 non-students}}$}

(b) Can 1200 vehicles bring in RM10000?

No. Even if all the cars were from non-students, the most you could get is  

1200 × 7.50 = RM9000

(c) Possible combinations for 1200 vehicles  

Revenue = 5.00x + 7.50y = 5.00x + 7.50(1200 -x) = 5.00x + 9000 - 7.50x =  

Revenue = 9000 - 2.50x

The maximum revenue of RM9000 occurs when there are no student cars and 1200 non-student cars.

For each student car that enters and displaces a non-student, the revenue drops by RM2.50.

Finally. when there are 1200 student cars and no non-students, the revenue has dropped to a minimum of RM6000.  

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