All categories

3 years ago

Solve the following equation. Then place the correct number in the box provided. x/9 = 1

Mathematics

2 answers:

Alchen [17]3 years ago

6 0

$\frac{x}{9} = 1 \\ \\ 1. \: x = 1 \times 9 \\ \\ 2. \: x = 9$

BARSIC [14]3 years ago

3 0

First, we need to isolate the variable x.

To isolate x, we simply need to multiply both sides by 9, which would result in x = 9

You might be interested in

Which expression is equivalent to 24 ⋅ 2−7? 1 over 2 to the power of 11 1 over 2 to the power of 3 23 211

lawyer [7]

I over 2 to the power 11

5 0

3 years ago

Read 2 more answers

11. Suppose a force F varies as the product of two masses, m and M, inversely as

almond37 [142]

Answer:

$k = 6.74 * 10^{-5}$

Step-by-step explanation:

Given

$F\ \alpha\ \frac{m * M}{r^2}$ --- The variation

$F = 675; m = 125; M = 225; r = 0.0530$

Required

Determine the constant of proportionality (k)

$F\ \alpha\ \frac{m * M}{r^2}$

Express as an equation

$F = k\frac{m * M}{r^2}$

Multiply both sides by $r^2$

$Fr^2 = km * M$

Make k the subject

$k = \frac{Fr^2}{m * M}$

Given that: $F = 675; m = 125; M = 225; r = 0.0530$

We have:

$k = \frac{675* 0.0530^2}{125 * 225}$

$k = \frac{1.896075}{28125}$

$k = 0.000067416$

This can be represented as:

$k = 6.74 * 10^{-5}$

5 0

3 years ago

Differentiate with respect to t

True [87]

Answer:

$\displaystyle 2x\frac{dx}{dt}-3y^2\frac{dy}{dt}+4z^3\frac{dz}{dt}=0$

Step-by-step explanation:

We want to differentiate the equation:

$x^2-y^3+z^4=1$

With respect to t, where x, y, and z are functions of t.

So:

$\displaystyle \frac{d}{dt}\left[x^2-y^3+z^4\right]=\frac{d}{dt}\left[1\right]$

Implicitly differentiate on the left. On the right, the derivative of a constant is simply zero. Hence:

$\displaystyle 2x\frac{dx}{dt}-3y^2\frac{dy}{dt}+4z^3\frac{dz}{dt}=0$

8 0

3 years ago

The length of a rectangle is increasing at a rate of 4 meters per day and the width is increasing at a rate of 1 meter per day.

puteri [66]

Answer:

$\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

Area of a Rectangle: A = lw

l is length
w is width

Calculus

Derivatives

Derivative Notation

Implicit Differentiation

Differentiation with respect to time

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

Step-by-step explanation:

Step 1: Define

$\displaystyle l = 10 \ meters$

$\displaystyle \frac{dl}{dt} = 4 \ m/day$

$\displaystyle w = 23 \ meters$

$\displaystyle \frac{dw}{dt} = 1 \ m/day$

Step 2: Differentiate

[Area of Rectangle] Product Rule: $\displaystyle \frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

Step 3: Solve

[Rate] Substitute in variables [Derivative]: $\displaystyle \frac{dA}{dt} = (10 \ m)(1 \ m/day) + (23 \ m)(4 \ m/day)$
[Rate] Multiply: $\displaystyle \frac{dA}{dt} = 10 \ m^2/day + 92 \ m^2/day$
[Rate] Add: $\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

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

Unit: Implicit Differentiation

Book: College Calculus 10e

8 0

3 years ago

Jesse and Larry entered a pie eating contest. Jesse ate 2 less

miss Akunina [59]

Answer:

Step-by-step explanation:

i did a test on it and got 100%

6 0

2 years ago