State whether the two triangles below are congruent? B А С Q S R

Which choice is equivalent to the quotient shown here when x > 0

Zinaida [17]

Answer:

this is equal to √(3x²/12) = √(x²/4) = x/2

Step-by-step explanation:

5 0

2 years ago

Find the value of x. (6x + 1) (9x - 23)

shtirl [24]

Answer:

Step-by-step explanation:

(6x+1)(9x-23)= 6x(9x-23)+1(9x-23)

= 54x²-138+9x-23

= 54x²+9x-161

3 0

2 years ago

Find the vertex of the graph of the function.

lapo4ka [179]

Answer: The correct options are (1) (5,10), (2) (3,-3), (3) x = -1, (4) $y=(x+2)^2+3$ , (5) 21s and (6) 0, -1, and 5.

Explanation:

Te standard form of the parabola is,

$f(x)=a(x-h)^2+k$ .....(1)

Where, (h,k) is the vertex of the parabola.

(1)

The given equation is,

$f(x)=(x-5)^2+10$

Comparing this equation with equation (1),we get,

$h=5$ and $k=10$

Therefore, the vertex of the graph is (5,10) and the fourth option is correct.

(2)

The given equation is,

$f(x)=3x^2-18x+24$

$f(x)=3(x^2-6x)+24$

To make perfect square add $(\frac{b}{2a})^2$ , i.e., $9$ . Since there is factor 3 outside the parentheses, so subtract three times of 9.

$f(x)=3(x^2-6x+9)+24-3\times 9$

$f(x)=3(x-3)^2-3$

Comparing this equation with equation (1),we get,

$h=3$ and $k=-3$

Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.

(3)

The given equation is

$f(x)=4x^2+8x+7$

$f(x)=4(x^2+2x)+7$

To make perfect square add $(\frac{b}{2a})^2$ , i.e., $1$ . Since there is factor 4 outside the parentheses, so subtract three times of 1.

$f(x)=4(x^2+2x+1)+7-4$

$f(x)=4(x+1)^2+3$

Comparing this equation with equation (1),we get,

$h=-1$ and $k=3$

The vertex of the equation is (-1,3) so the axis is x=-1 and the correct option is 2.

(4)

The given equation is,

$y=x^2+4x+7$

To make perfect square add $(\frac{b}{2a})^2$ , i.e., $2^2$ .

$f(x)=x^2+4x+4+7-4$

$f(x)=(x+2)^2+3$

Therefore, the correct option is 4.

(5)

The given equation is,

$h=-16t^2+672t$

It can be written as,

$h=-16(t^2-42t)$

It is a downward parabola. so the maximum height of the function is on its vertex.

The x coordinate of the vertex is,

$x=\frac{b}{2a}$

$x=\frac{42}{2}$

$x=21$

Therefore, after 21 seconds the projectile reach its maximum height and the correct option is first.

(6)

The given equation is,

$f(x)=3x^3-12x^2-15x$

$f(x)=3x(x^2-4x-5)$

Use factoring method to find the factors of the equation.

$f(x)=3x(x^2-5x+x-5)$

$f(x)=3x(x(x-5)+1(x-5))$

$f(x)=3x(x-5)(x+1)$

Equate each factor equal to 0.

$x=0,-1,5$

Therefore, the zeros of the given equation is 0, -1, 5 and the correct option is 2.

3 0

3 years ago

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: $y''(-1,-2) = \frac{-4}{3}$