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3 years ago

If g=26.4 and F=35 find h

Mathematics

2 answers:

frutty [35]3 years ago

6 0

Can you tell us the equation please?

Free_Kalibri [48]3 years ago

3 0

I believe its 43.6 too

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What is the dilation for the transformation below?

lys-0071 [83]

Answer:

The dialation is of 2

Step-by-step explanation:

QR = 3 units

QS = 2 units

RS = 2 units

Q'R' = 6 units

Q'S' = 4 units

R'S' = 4 units

So as you can see, this triangle has doubled it's size, since it has been multiplied by 2.

Hope this helped!

Have a supercalifragilisticexpialidocious day!

4 0

3 years ago

Please help!!!!!!! How many inches would a point on the outer edge of the large gear travel in a 150º rotation? (The gear has a

damaskus [11]

In a full rotation, 360°, a point on the edge of a gear would travel a distance equal to its circumference. The circumference of a circle is:

C=2πr, and since we are only traveling 150°, we need to set up an appropriate ratio for circumference to the distance the point travels:

d/C=150/360

d=5C/12 and since C=2πr

d=10πr/12

d=5πr/12, and since r=4in

d=5*4π/12 in

d=20π/12 in

d=5π/3 in

d≈5.24 in (to nearest hundredth of an inch)

4 0

3 years ago

Find x. please help with this question

kiruha [24]

Answer:

11.875

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

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

-19x + 23x = 44 what is x

matrenka [14]

Step-by-step explanation:

- 19 x + 23 x = 44

4x = 44

x = 44/ 4

x = 11

Hope it will help :)

6 0

3 years ago

Read 2 more answers