Equation to 2 cups peaches to 3 cups cherries to make 60 cups how many cups of peaches would i need

You want to convert 50°F to degrees Celsius. Solve the formula for C, and then use it to convert the temperature. What is the co

Zepler [3.9K]

9514 1404 393

Answer:

C = (5/9)(F -32)
50°F ⇒ 10°C

Step-by-step explanation:

Starting with the formula for degrees Fahrenheit, we can solve for degrees Celsius.

F = 32 +9/5C

F -32 = (9/5)C . . . . . subtract 32

(5/9)(F -32) = C . . . . multiply by the inverse of the coefficient of C

The formula for Celsius temperature from Fahrenheit temperature is ...

C = (5/9)(F -32)

__

For F = 50, the corresponding Celsius temperature is ...

C = (5/9)(50 -32) = (5/9)(18) = 10 . . . degrees Celsius

5 0

3 years ago

Which of the following illustrates the truth value of the given conditional statement? If dogs have five legs, then cats have th

jeka94

Answer: In this conditional statement, all of the results are F.

There is no relationship between dogs having five legs and cats having three tails. Even if you took the converse and said "If cats have three tails, then dogs have five legs", there is no truth in that statement.

4 0

3 years ago

Read 2 more answers

Which question is being asked in this story problem?

olya-2409 [2.1K]

The answer is C.

The question asks for the total amount of tires on all of the toy tractors.

7 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

Round 76.625 to the nearest hundredths

lidiya [134]

Is 76.63 because:
If it is 5 or above, you give it one more point .
If it is 4 or below, you let it stay still.

4 0

3 years ago