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tatyana61 [14]
3 years ago
12

What are the solutions to the quadratic equation x^2-16=0

Mathematics
1 answer:
maks197457 [2]3 years ago
4 0
Good evening.


This is a incomplete quadratic equation, because it does not have the term bx. Therefore, we can solve this faster with the following strategy:

\mathsf{x^2 - 16 = 0}

Add 16 to both sides:

\mathsf{x^2 - 16 + 16 = 0 + 16}\\ \\ \mathsf{x^2 = 16}

Take the square root:

\mathsf{\sqrt {x^2} = \pm\sqrt{16}}\\ \\ \mathsf{\boxed{\mathsf{x = \pm4} }}

Therefore:

\boxed{\boxed{\mathsf{S=\{-4, \ +4\}}}}
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A triangle has side lengths of (q+r)(q+r) centimeters, (5q-10s)(5q−10s) centimeters, and (5s-7r)(5s−7r) centimeters. Which expre
Stolb23 [73]

The expression represents the perimeter, in centimeters, of the triangle is 6q - 6r - 5s

<h3>What is the perimeter?</h3>

The formula for perimeter of a triangle is expressed as;

Perimeter = a + b + c

Where a , b and c are the lengths of its side

  • (q+r)
  • (5q-10s)
  • (5s-7r)

Now, let's substitute the values

Perimeter = (q + r) + (5q - 10s) + (5s - 7r)

expand the bracket

Perimeter = q + r + 5q - 10s + 5s - 7r

collect like terms

Perimeter = q + 5q + r - 7r -10s + 5s

Add like terms

Perimeter = 6q - 6r - 5s

Thus, the expression represents the perimeter, in centimeters, of the triangle is 6q - 6r - 5s

Learn more about perimeter here:

brainly.com/question/24571594

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1 year ago
DUE IN 5 MINS PLEASE HELP
Maksim231197 [3]

Answer:

12

Step-by-step explanation:

you can pack one pack in 20 min

there are three 20 min in one hour

multiply three times the four hours and you get 12

5 0
2 years ago
I need help on this question <br> Factor by grouping.<br> 2u^3+3u^2+14u+21
Andru [333]
U^2(2u+3)+7(2u+3)

(u^2+7)(2u+3)
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2 years ago
Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

7 0
3 years ago
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miss Akunina [59]

Answer:

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8 0
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