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3 years ago

What two numbers have the product of 234 and the sum of 31

1 answer:

8 0

The answer is 13 and 18

Two numbers have the product of 234: x * y = 234
Two numbers have the sum of 31: x + y = 31

This is the system of equations:
x * y = 234
x + y = 31
_______
x * y = 234
y = 31 - x
_______
x(31 - x) = 234
31x - x² = 234 (-1)
-31x + x² = -234
x² - 31x = -234
x² - 31x + 234 = 0

According to the quadratic equation formula:
$x_{1,2} = \frac{-b+/- \sqrt{ b^{2}-4ac } }{2a} = \frac{-(31)+/- \sqrt{ (-31)^{2}-4*1*234 } }{2*1} = \\ \\ = \frac{31+/- \sqrt{ 961-936 } }{2} =\frac{31+/- \sqrt{25} }{2} = \frac{31+/-5}{2} \\ \\ x_1 =\frac{31+5}{2}= \frac{36}{2}=18 \\ \\ x_1 =\frac{31-5}{2}= \frac{26}{2}=13$

When: x1 = 18, y1 = 31 - 18 = 13
When: x2 = 13, y1 = 31 - 13 = 18

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Hi, can anyone help with b) and c)?

VARVARA [1.3K]

Answer:

steps below

Step-by-step explanation:

From the graph: parabola pass (0,4) (3,4) (1,2)

equation: y = ax² + bx + c

(0,4): c = 4

(3,4): 9a + 3b + 4 = 4 9a+3b = 0 3a + b = 0 ... (1)

(1,2): a + b + 4 = 2 a+b = -2 ... (2)

(1)-(2): 2a = 2 a = 1

(2): b = -3

parabola equation: f(x) = x² - 3x +4

(a) f(3) = 3² - 3*3 + 4 = 4

(b) normal line: parallel to tangent line y = 3x -5 at (3,4)

slope of line N: -1/3 pass (3,4)

equation of N: (y-4)/(x-3) = -1/3 y-4 = -1/3 x + 1

line N: 1/3 x + y - 5 = 0

8 0

3 years ago

Which unit price is the highest?

Leviafan [203]

Answer:peppers

Step-by-step explanation:

4 0

3 years ago

Please help me to prove this!

Ymorist [56]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → A + B = π - C

→ B + C = π - A

→ C + A = π - B

→ C = π - (B + C)

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B

Use the Double Angle Identity: sin 2A = 2 sin A · cos A

Use the Cofunction Identity: cos (π/2 - A) = sin A

Proof LHS → Middle:

$\text{LHS:}\qquad \qquad \cos \bigg(\dfrac{A}{2}\bigg)+\cos \bigg(\dfrac{B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)$

$\text{Sum to Product:}\qquad 2\cos \bigg(\dfrac{\frac{A}{2}+\frac{B}{2}}{2}\bigg)\cdot \cos \bigg(\dfrac{\frac{A}{2}-\frac{B}{2}}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)$

$\text{Given:}\qquad \quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)$

$\text{Sum/Difference:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)$

$\text{Double Angle:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{2(A+B)}{2(2)}\bigg)\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+2\sin \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)$

$\text{Factor:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]$

$\text{Cofunction:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg]$

$\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)$

$\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{\pi -C}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)$

LHS = Middle $\checkmark$

Proof Middle → RHS:

$\text{Middle:}\qquad 4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)\\\\\\\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)$

Middle = RHS $\checkmark$

3 0

3 years ago

PLEASE HELP ASAP GIVING BRAINLIEST

Lady_Fox [76]

Answer:

Part 1) $y=32.5x+11.95$

Par 2) see the explanation

Part 3) $\$401.95$

Step-by-step explanation:

Part 1) Write an equation in slope intercept form to represent this problem

Let

x ----> the number of tickets

y ---> the total cost

The linear equation in slope intercept form is equal to

$y=mx+b$

where

m is the slope or unit rate

b is the y-intercept (one-time processing fee)

In this problem we have

$b=\$11.95$

ordered pairs (4,141.95) and (7,239.45)

Find the slope

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values in the formula

$m=\frac{239.45-141.95}{7-4}$

$m=\frac{97.5}{3}$

$m=32.5$

substitute the values in the linear equation

$y=32.5x+11.95$

Part 2) What does your slope represent in this problem? What does your y-intercept represent in this problem?

a) The slope of the linear equation represent the unit rate, so the slope represents the cost of one ticket

$m=\$32.5\ per\ ticket$

b) The y-intercept is the value of y when the value of x is equal to zero

In this context, the y-intercept represent the one-time processing fee

$b=\$11.95$

Part 3) Use the equation to find how much Alan would have to pay for 12 tickets

For x=12 tickets

substitute in the linear equation

$y=32.5(12)+11.95$

$y=\$401.95$

3 0

4 years ago

Solve for x and y x+1/2 - y+4/11=2 ; x+3/2 + 2y+3/17=5

puteri [66]

The value of x and y is 5 , 7

<h3>What is an Equation ?</h3>

An expression consists of variables , constants and mathematical operators , when equated by any other algebraic expression or a constant it becomes an equation.

The equation given in the question is

(x + 1)/2 -(y+4)/11 = 2

(x+3) /2 + (2y+3)/17 = 5

It has to be solved for x and y ,

The simplification of the equation has to be done

and the equation is

11x-2y = 41 -----------1

and

17x +4y = 113 -------------------2

to find the solution elimination method will be followed as

Multiplying equation 1 by 2 and adding the both equations

39x = 195

x = 5

Substituting this value in equation 1

55 -2y = 41

y = 7

The value of x and y is 5 , 7

To know more about Equation

brainly.com/question/10413253

#SPJ1

3 0

2 years ago