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2 years ago

What are the solutions for x to the equation (x-3)(x+8)=0

Mathematics

1 answer:

hichkok12 [17]2 years ago

7 0

For this case we have a factored expression, of a second degree equation:

( $(x-3) (x + 8) = 0$

To find the solutions:

$x-3 = 0$

we add 3 to both sides of the equation:

$x=3$

$x+8=0$

We subtract 8 on both sides of the equation:

$x = -8$

Thus, the solutions of the equation are:

$x_ {1} = 3\\x_ {2} = - 8$

Answer:

$x_ {1} = 3\\x_ {2} = - 8$

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Two semicircles are attached to the sides of a rectangle as shown.

NeTakaya

Answer:

157 in²

Step-by-step explanation:

The area of the shape is the sum of three sections

1. Rectangle

A = 5*14 = 70 in²

2. Bigger semicircle

A = 1/2π(14/2)² = 76.96 ≈ 77 in² (rounded)

3. Smaller semicircle

A = 1/2π(5/2)² = 9.8174 ≈ 10 in² (rounded)

Total area:

70 + 77 + 10 = 157 in²

4 0

2 years ago

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Someone help me with number 1 and 2 please

adell [148]

1 is 25
2 is 7.42
if you need explanation please comment and let me know

7 0

2 years ago

Help help help please

Olin [163]

Answer:

Step-by-step explanation:

a² + b² = c²

a² + 9² = 12²

a² + 81 = 144

a² = 63

a = √63

4 0

2 years ago

Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''

Nonamiya [84]

Answer:

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

$D^2-1=0$

$(D-1)(D+1)=0$

D=1,-1

Then , the solution of given differential equation

$y=e^x,y=e^{-x}$

2. $7x^2y''+14xy'-14 y=0$

Y= $y=x^3$

$y=3x^2$

$y''=6x$

Substitute in the given differential equation

$42x^3+42x^3-14x^3\neq 0$

Hence, $x^3$ is not a solution of given differential equation

$e^x,e^{-x}$ are also not a solution of given differential equation.

y= $x^{-2}$

$y'=-2x^{-3}$

$y''=6x^{-4}$

Substitute the values in the differential equation

$7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}$

= $42x^{-2}-28x^{-2}-14x^{-2}=0$

Hence, $x^{-2}$ is a solution of given differential equation.

c. $7x^2y''-42y=0$

$y=x^3$

$y'=3x^2$

$y''=6x$

Substitute the values in the differential equation

$42x^3-42x^3=0$

Hence, $x^3$ is a solution of given differential equation.

a- $e^x,e^{-x}$

b- $x^{-2}$

c- $x^3$

6 0

2 years ago

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Prove the identity (cosx+cosy)^2+(sinx-siny)^2= 2+2cos(x+y)

astra-53 [7]

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1 + cos^2(y) + 2cos(x)*cos(y) - 2*sin(x)*cos(y) + sin^2(y)
These two in bold also make 1

2 + 2cos(x)*cos(y) - 2*sin(X)*sin(y) Bring out a common factor of 2
2 +2(cos(x)*cos(y) - sin(x)*sin(y) )

but cos(x+y ) = cos(x)*cos(y) - sin(x)*sin(y)

2 + 2* cos(x + y) is your final answer.

4 0

2 years ago

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