Need help asap!!!! What’s the answer to this?? Having any answers

Tony bought x cans of tennis balls for $3 each. He bought the same number of water bottles for $5. He also

NeX [460]

What are you asking exactly?

7 0

3 years ago

Six out of 10 kids prefer swimming to hiking. How many kids out of 100 prefer swimming?

eimsori [14]

6/10 = 0.6

0.6 × 100 = 60

Answer = B.60

~Aamira~

~Hope this helped~

8 0

3 years ago

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The sets A, B, and C are defined as follows:

PilotLPTM [1.2K]

The element of B × A × C is equivalent to the element of A × B × C.

B × A × C = {tall foam nonfat, tall foam whole, tall no-foam nonfat, tall no-foam whole, grande foam non fat, grande foam whole, grande no foam nonfat, grande no foam whole, venti foam nonfat,, venti foam whole, venti nofoam nonfat, venti nofoam whole}

For the element of B × C

B × C = {foam nonfat, foam whole, no foam nonfat, no foam whole}

<h3>Product of sets</h3>

Given the following sets

A = {tall, grande, venti}

B = {foam, no-foam}

C = {non-fat, whole}

The set of A * B is expressed as:

A * B = {tall foam, tall no foam, grande foam, grande no foam, venti foam, venti nofoam}

A × B × C = {tall foam nonfat, tall foam whole, tall no-foam nonfat, tall no-foam whole, grande foam non fat, grande foam whole, grande no foam nonfat, grande no foam whole, venti foam nonfat,, venti foam whole, venti nofoam nonfat, venti nofoam whole}

The element of B × A × C is equivalent to the element of A × B × C.

B × A × C = {tall foam nonfat, tall foam whole, tall no-foam nonfat, tall no-foam whole, grande foam non fat, grande foam whole, grande no foam nonfat, grande no foam whole, venti foam nonfat,, venti foam whole, venti nofoam nonfat, venti nofoam whole}

For the element of B × C

B × C = {foam nonfat, foam whole, no foam nonfat, no foam whole}

Learn more on set notation here: brainly.com/question/24462379

8 0

2 years ago

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$

Use the Basic Power Rule:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')$

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]$

Simplifying it, we have:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]$

We can rewrite the 2nd derivative using exponential rules:

$\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}$

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}$

When we evaluate this using order of operations, we should obtain our answer:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

5 0

3 years ago

Convert 35 pounds of candy to stones. (1 stone = 14 pounds)

bazaltina [42]

Answer:

2.4 stone

Step-by-step explanation: i think i did it in my head

7 0

3 years ago

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