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3 years ago

Please show the answer with the work, GodBless.

Mathematics

1 answer:

Yanka [14]3 years ago

8 0

Well, we know that y=1/2x+5 from the first equation. if we substitute that into the second equation, it becomes 2x + 1/2x +5 =1. solve for x. 2x+1/2x= -4, 2 1/2x=-4, x=-1.6. plug that in to one of the equations. 2(-1.6)+y=1, -3.2+y=1, y=-2.2. the solution is (-1.6, -2.2)

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If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form

algol13

Answer:

$\frac{d^2y}{dx^2} = \frac{-4}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Basic Power Rule]: $-y'-6x^2=2yy'$
[Algebra] Isolate y' terms: $-6x^2=2yy'+y'$
[Algebra] Factor y': $-6x^2=y'(2y+1)$
[Algebra] Isolate y': $\frac{-6x^2}{(2y+1)}=y'$
[Algebra] Rewrite: $y' = \frac{-6x^2}{(2y+1)}$

Step 3: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}$
[Derivative] Back-Substitute y': $y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}$

Step 4: Find Slope at Given Point

[Algebra] Substitute in x and y: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Multiply: $y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}$
[Pre-Algebra] Divide: $y''(-1,-2) = \frac{-36+24 }{9}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-12}{9}$
[Pre-Algebra] Simplify: $y''(-1,-2) = \frac{-4}{3}$

6 0

2 years ago

Question Down Below...

Delicious77 [7]

5/12 equals to 20/48.

48÷12=4
5×4=20

7 0

2 years ago

Read 2 more answers

E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis .Find the coordinates of H and G. Please need answe

timurjin [86]

Answer:

coordinates of H = (1, 0)

Coordinates of G = ( - 3.6, -2) or (5.6, -2)

Step-by-step explanation:

E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis.

Let the coordinates of H is (x, 0) and G is (a, b).

The length of side EF is

$EF = \sqrt{(5 -2)^2 + (3 +1)^2} = 5$

So,

$EH = \sqrt{(5 -x)^2 + (3 -0)^2} = 5\\\\(5 -x)^2+ 9 = 25\\\\5 - x = 4\\\\x = 1$

And

$GH = \sqrt{(a -x)^2+ b^2} = 5\\\\(a -1)^2+ b^2 = 25\\\\a^2 + b^2 + 1 - 2 a = 25\\\\a^2 + b^2 - 2a = 24 .... (1)$

Now

$FG = \sqrt{(a -2)^2+ (b +1)^2} = 5\\\\(a -2)^2+ (b+1)^2 = 25\\\\a^2 + b^2 + 2b - 2 a = 20\\ ..... (2)$

Solving (1) and (2)

b = - 2 ,

$a = \frac{2 \pm\sqrt{4 + 80}}{2}\\\\a = \frac{2 \pm 9.2}{2}\\\\a = - 3.6, 5.6$

3 0

3 years ago

Maria wrote the equation log(x/2)+log(20/x^2)=log8. What is the solution to Maria's equation?

Fed [463]

The solution is x=5/4.

We use the properties of logs to rewrite the equation:
$\log[(\frac{x}{2})(\frac{20}{x^2})]=\log8
\\
\\\log(\frac{20x}{2x^2})=\log8
\\
\\\log(\frac{10}{x})=\log8$

Get all of the logs on the same side of the equation y subtracting log 8:
$\log(\frac{10}{x})-\log8=\log8 - \log8
\\
\\\log(\frac{10}{x})-\log8=0$

Use the properties of logs to rewrite:
$\log(\frac{10}{x}\div8)=0
\\
\\\log(\frac{10}{x}\div\frac{8}{1})=0
\\
\\\log(\frac{10}{x}\times\frac{1}{8})=0
\\
\\\log(\frac{10}{8x})=0$

Exponentiate:
$10^0=\frac{10}{8x}
\\
\\1=\frac{10}{8x}$

Multiply both sides by 8x:
1*8x = (10/8x)*8x
8x=10

Divide both sides by 8:
8x/8 = 10/8
x = 10/8 = 5/4

3 0

3 years ago

Read 2 more answers

Please explain how to begin to solve this

Sergio039 [100]

Answer:

6.00

Step-by-step explanation:

Use Pythagorean theorem:

c² = a² + b²

9.22² = a² + 7²

85.0 = a² + 49

a² = 36.0

a = 6.00

7 0

3 years ago