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3 years ago

A rectangular living room has a perimeter of 38 meters and an area of 90 square meters.

Mathematics

1 answer:

ruslelena [56]3 years ago

3 0

Answer:

10 meters by 9 meters

Step-by-step explanation:

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Plsss help quick three question

yKpoI14uk [10]

Answer:

1) A

2) min; min; max; max

3) y = x² + 5x - 3

Step-by-step explanation:

f(x) = x² + 2(x)(5) + 5² - 5² + 24

f(x) = (x + 5)² - 25 + 24

f(x) = (x + 5)² - 1

In ax² + bx + c,

if a > 0, it's a min

if a < 0, it's a max

y = ax² + bx + c

Using (0,-3)

-3 = a(0)² + b(0) + c

c = -3

y = ax² + bx - 3

Using (1,3)

3 = a + b - 3

a + b = 6

Using (-1,-7)

-7 = a(-1)² + b(-1) - 3

-7 + 3 = a - b

a - b = -4

b = a + 4

a + (a + 4) = 6

2a = 2

a = 1

b = 5

y = x² + 5x - 3

7 0

2 years ago

How to find the base of an isosceles when given one side

MA_775_DIABLO [31]

Double the side and subtract by 360

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2 years ago

Please Help Me On These Math Questions!!!!! Solve Each System By Elimination

nekit [7.7K]

11) -x + y = -1 ; 2x - y = 0
y = -1 + x ; 2x - (-1+x) = 0 ⇒ 2x + 1 - x = 0 ⇒x = -1
y = -1 + (-1) ⇒ y = -2

12) -2x + y = -20 ; 2x + y = 48
y = -20 + 2x ; 2x + (-20 + 2x) = 48 ⇒ 2x -20 + 2x = 48 ⇒ 4x = 48 + 20
4x = 68 ⇒ x = 68/4 ⇒ x = 17
y = -20 + 2(17) ⇒ y = -20 + 34 ⇒ y = 14

13) 3x -y = -2 ; -2x + y = 3
y = 3 + 2x ; 3x - (3 + 2x) = -2 ⇒ 3x - 3 - 2x = -2 ⇒ x = -2 + 3 ⇒ x = 1
y = 3 + 2(1) ⇒ y = 3 + 2 ⇒ y = 5

14) x - y = 4 ; x - 2y = 10
x = 4 + y ; (4 + y) - 2y = 10 ⇒ 4 + y - 2y = 10 ⇒ 4 - y = 10
⇒ -y = 10 - 4 ⇒ -y = 6 ⇒ y = -6
x = 4 + (-6) ⇒ x = 4 - 6 ⇒ x = -2

15) x + 2y = 5 ; 3x + 2y = 17
x = 5 - 2y ; 3(5-2y) + 2y = 17 ⇒ 15 - 6y + 2y = 17 ⇒ -4y = 17 - 15
⇒ -4y = 2 ⇒ y = -2/4 ⇒ y = -1/2
x = 5 - 2(-1/2) ⇒ x = 5 + 2/2 ⇒ x = 5 + 1 ⇒ x = 6

5 0

2 years ago

Read 2 more answers

Solve the following equation by factoring: 9x^2 – 3x – 2 = 0

crimeas [40]

The answer is A. X=-1/3 or x=2/3

5 0

3 years ago

Read 2 more answers

Find all solutions to the following quadratic equations, and write each equation in factored form.

dexar [7]

Answer:

(a) The solutions are: $x=5i,\:x=-5i$

(b) The solutions are: $x=3i,\:x=-3i$

(c) The solutions are: $x=i-2,\:x=-i-2$

(d) The solutions are: $x=-\frac{3}{2}+i\frac{\sqrt{7}}{2},\:x=-\frac{3}{2}-i\frac{\sqrt{7}}{2}$

(e) The solutions are: $x=1,\:x=-1,\:x=\sqrt{5}i,\:x=-\sqrt{5}i$

(f) The solutions are: $x=1$

(g) The solutions are: $x=0,\:x=1,\:x=-2$

(h) The solutions are: $x=2,\:x=2i,\:x=-2i$

Step-by-step explanation:

To find the solutions of these quadratic equations you must:

(a) For $x^2+25=0$

$\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\x^2+25-25=0-25$

$\mathrm{Simplify}\\x^2=-25$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-25},\:x=-\sqrt{-25}$

$\mathrm{Simplify}\:\sqrt{-25}\\\\\mathrm{Apply\:radical\:rule}:\quad \sqrt{-a}=\sqrt{-1}\sqrt{a}\\\\\sqrt{-25}=\sqrt{-1}\sqrt{25}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\\sqrt{-25}=\sqrt{25}i\\\\\sqrt{-25}=5i$

$-\sqrt{-25}=-5i$

The solutions are: $x=5i,\:x=-5i$

(b) For $-x^2-16=-7$

$-x^2-16+16=-7+16\\-x^2=9\\\frac{-x^2}{-1}=\frac{9}{-1}\\x^2=-9\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\x=\sqrt{-9},\:x=-\sqrt{-9}$

The solutions are: $x=3i,\:x=-3i$

(c) For $\left(x+2\right)^2+1=0$

$\left(x+2\right)^2+1-1=0-1\\\left(x+2\right)^2=-1\\\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x+2=\sqrt{-1}\\x+2=i\\x=i-2\\\\x+2=-\sqrt{-1}\\x+2=-i\\x=-i-2$

The solutions are: $x=i-2,\:x=-i-2$

(d) For $\left(x+2\right)^2=x$

$\mathrm{Expand\:}\left(x+2\right)^2= x^2+4x+4$

$x^2+4x+4=x\\x^2+4x+4-x=x-x\\x^2+3x+4=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are:

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=3,\:c=4:\quad x_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}$

$x_1=\frac{-3+\sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}=\quad -\frac{3}{2}+i\frac{\sqrt{7}}{2}\\\\x_2=\frac{-3-\sqrt{3^2-4\cdot \:1\cdot \:4}}{2\cdot \:1}=\quad -\frac{3}{2}-i\frac{\sqrt{7}}{2}$

The solutions are: $x=-\frac{3}{2}+i\frac{\sqrt{7}}{2},\:x=-\frac{3}{2}-i\frac{\sqrt{7}}{2}$

(e) For $\left(x^2+1\right)^2+2\left(x^2+1\right)-8=0$

$\left(x^2+1\right)^2= x^4+2x^2+1\\\\2\left(x^2+1\right)= 2x^2+2\\\\x^4+2x^2+1+2x^2+2-8\\x^4+4x^2-5$

$\mathrm{Rewrite\:the\:equation\:with\:}u=x^2\mathrm{\:and\:}u^2=x^4\\u^2+4u-5=0\\\\\mathrm{Solve\:with\:the\:quadratic\:equation}\:u^2+4u-5=0$

$u_1=\frac{-4+\sqrt{4^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}=\quad 1\\\\u_2=\frac{-4-\sqrt{4^2-4\cdot \:1\left(-5\right)}}{2\cdot \:1}=\quad -5$

$\mathrm{Substitute\:back}\:u=x^2,\:\mathrm{solve\:for}\:x\\\\\mathrm{Solve\:}\:x^2=1=\quad x=1,\:x=-1\\\\\mathrm{Solve\:}\:x^2=-5=\quad x=\sqrt{5}i,\:x=-\sqrt{5}i$

The solutions are: $x=1,\:x=-1,\:x=\sqrt{5}i,\:x=-\sqrt{5}i$

(f) For $\left(2x-1\right)^2=\left(x+1\right)^2-3$

$\left(2x-1\right)^2=\quad 4x^2-4x+1\\\left(x+1\right)^2-3=\quad x^2+2x-2\\\\4x^2-4x+1=x^2+2x-2\\4x^2-4x+1+2=x^2+2x-2+2\\4x^2-4x+3=x^2+2x\\4x^2-4x+3-2x=x^2+2x-2x\\4x^2-6x+3=x^2\\4x^2-6x+3-x^2=x^2-x^2\\3x^2-6x+3=0$

$\mathrm{For\:}\quad a=3,\:b=-6,\:c=3:\quad x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:3\cdot \:3}}{2\cdot \:3}\\\\x_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{0}}{2\cdot \:3}\\x=\frac{-\left(-6\right)}{2\cdot \:3}\\x=1$

The solutions are: $x=1$

(g) For $x^3+x^2-2x=0$

$x^3+x^2-2x=x\left(x^2+x-2\right)\\\\x^2+x-2:\quad \left(x-1\right)\left(x+2\right)\\\\x^3+x^2-2x=x\left(x-1\right)\left(x+2\right)=0$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x=0\\x-1=0:\quad x=1\\x+2=0:\quad x=-2$

The solutions are: $x=0,\:x=1,\:x=-2$

(h) For $x^3-2x^2+4x-8=0$

$x^3-2x^2+4x-8=\left(x^3-2x^2\right)+\left(4x-8\right)\\x^3-2x^2+4x-8=x^2\left(x-2\right)+4\left(x-2\right)\\x^3-2x^2+4x-8=\left(x-2\right)\left(x^2+4\right)$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x-2=0:\quad x=2\\x^2+4=0:\quad x=2i,\:x=-2i$

The solutions are: $x=2,\:x=2i,\:x=-2i$

3 0

2 years ago