Completing the square

x^2-5x-14

Answers

Answer 1

Answer: completing the square is normally used when the equation is equal to zero
]
if we were to factor, we would get (what 2 number multiply to -14 and add to -5 then answer is -7 and 2)
(x+2)(x-7)

if this was set to zero then the number that would work are -2 and 7

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3+3×3-3+3=What's the answerHow do you get the answer
If RB = 5, then AB = ? a. 2.5 b. 5 c. 10

If f(x) = x2 and g(x) = 3x + 1, then g(f(2))= ?

Answers

Answer:

Step-by-step explanation:

f(2) = 2² = 4

g(4) = 3(4) + 1 which is 13

Absolute value of -3.47 ?

Answers

Answer: 3.47

Step-by-step explanation:

Absolute value is distance from zero.

Which pair of triangles can be proven congruent by SAS?

Answers

Where The Whole Problem At. ?

Q6.2
(8 + 2 x 6)2 =

the two at the end of the brackets is squared it a small two

Answers

Answer: 1600

Step-by-step explanation:

2(8+2x6)

2(8+12)

2(20)

$40^(2)$

1600

Mark me as the brainiest answer.

(8+2x6)2^2
= (8+12)2^2
=(20)2^2
=20x4
=80

A decorative steel frame is shown below it is made of a circular section and and 6 stems the length of each stem is 45 cm work out the total length of Steel used in the frame

Answers

A total of 552.86 cm of steel is used in the frame development.

What is Equation Modelling?

Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem.

We have a decorative steel frame is shown below it is made of a circular section and and 6 stems the length of each stem is 45 cm.

Assume that the total length of the steel used is [x] cm. Then -

x = 2πr + 6r

x = 2 x 22/7 x 45 + 6 x 45

x = 282.86 + 270

x = 552.86 cm

Therefore, a total of 552.86 cm of steel is used in the frame development.

To solve more questions on circumferance, visit the link below-

brainly.com/question/15529702

#SPJ5

Answer:

552.78

Step-by-step explanation:

A rancher has 200 feet of fencing to enclose two adjacent corralsa.what dimensions should be used so that the enclosed area will be maximum b)what is the maximum area?

Answers

Answer:

a) Each corral should be 33⅓ ft long and 25 ft wide

b) The total enclosed area is 1666⅔ ft²

Step-by-step explanation:

I assume that the corrals have identical dimensions and are to be fenced as in the diagram below

Let x = one dimension of a corral

and y = the other dimension

(a) Dimensions to maximize the area

The total length of fencing used is:

4x + 3y = 200

4x = 200 – 3y

x = 50 - ¾y

The area of one corral is A = xy, so the area of the two corrals is

A = 2xy

Substitute the value of x

A = 2(50 - ¾y)y

A = 100 y – (³/₂)y²

This is the equation for a downward-pointing parabola:

A = (-³/₂)y² + 100y

a = -³/₂; b = 100; c = 0

The vertex (maximum) occurs at

y = -b/(2a) = 100 ÷ (2×³/₂) = 100 ÷ 3 = 33⅓ ft

4x + 3y = 100

Substitute the value of y

4x + 3(33⅓) = 200

4x + 100 = 200

4x = 100

x = 25 ft

Each corral should measure 33⅓ ft long and 25 ft wide.

Step 2. Calculate the total enclosed area

The enclosed area is 50 ft long and 33⅓ ft wide.

A = lw = 50 × 100/3 = 5000/3 = 1666⅔ ft²

Final answer:

The maximum area is achieved when the shared fence is 50 feet and the other two sides are 75 feet each, yielding a maximum area of 3750 square feet.

Explanation:

This problem can be solved by the principles of calculus. Assuming that the two corrals share a common side, we can say the total length of fencing is divided into two lengths (x and y). The optimization problem can be formed as follows:

x = length of the common fence
y = length of the other sides

Since the total length available is 200 feet, 2y + x = 200. The area A = xy. Substitute y=(200-x)/2 into the area formula to get a quadratic A = x(200-x)/2. This graph opens downwards, meaning the vertex is the maximum point. The x-coordinate of the vertex of a quadratic given in standard form like Ax^2 + Bx + C is -B/2A. Therefore, x = -B/2A = 200/(2*2) = 50. Substitute x back into y = (200-2x)/2 to get y = 75. So, the maximum area is achieved with a common side of 50 feet and the other sides being 75 feet each.

The maximum area A can be found by substituying these values back into the area formula: A = 75*50 = 3750 square feet.