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Karo-lina-s [1.5K]

3 years ago

6

I need to write an expression of this phrase: The amount of money Nina earned if she mowed the lawn for fifteen dollars and walk

ed
to dog for four hours?

1 answer:

Andrei [34K]3 years ago

7 0

Answer:

X = 15+4Y

Step-by-step explanation:

The amount of money we can set as X

Since we dont know the amount of money she gets each hour we can set that to Y

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Can you guys help me!!!

Gnesinka [82]

I cant see the question, id help if i could

7 0

3 years ago

50 POINTS !!<br><br><br> PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.

seraphim [82]

Answer:

73

Step-by-step explanation:

48^2 + 55^2 = 5329

square root of 5329 is 73

7 0

3 years ago

Read 2 more answers

Find the area of the region enclosed by the graphs of the functions

Vaselesa [24]

Answer:

$\displaystyle A = \frac{8}{21}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

Terms/Coefficients
Functions
Function Notation
Graphing
Solving systems of equations

<u>Calculus</u>

Area - Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

*Note:

<em>Remember that for the Area of a Region, it is top function minus bottom function.</em>

<u />

<u>Step 1: Define</u>

f(x) = x²

g(x) = x⁶

Bounded (Partitioned) by x-axis

<u>Step 2: Identify Bounds of Integration</u>

<em>Find where the functions intersect (x-values) to determine the bounds of integration.</em>

Simply graph the functions to see where the functions intersect (See Graph Attachment).

Interval: [-1, 1]

Lower bound: -1

Upper Bound: 1

<u>Step 3: Find Area of Region</u>

<em>Integration</em>

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^1_{-1} {[x^2 - x^6]} \, dx$
[Area] Rewrite [Integration Property - Subtraction]: $\displaystyle A = \int\limits^1_{-1} {x^2} \, dx - \int\limits^1_{-1} {x^6} \, dx$
[Area] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = \frac{x^3}{3} \bigg| \limit^1_{-1} - \frac{x^7}{7} \bigg| \limit^1_{-1}$
[Area] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = \frac{2}{3} - \frac{2}{7}$
[Area] Subtract: $\displaystyle A = \frac{8}{21}$

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

Unit: Area Under the Curve - Area of a Region (Integration)

Book: College Calculus 10e

6 0

3 years ago

IRISSAK [1]

Step-by-step explanation:

$a + b = 6$

$ab = 1 \times 5 = 5$

$a = 1 \: \: \: \: \: \: \: \: b = 5$

$( {x}^{2} + x) + (5x + 5)$

$x(x + 1) + 5(x + 1)$

$(x + 1)(x + 5)$

Hope this is correct and helpful

HAVE A GOOD DAY!

8 0

4 years ago

Is -2 a solution to 4x^(3)-x^(2)+16x-4

sergiy2304 [10]

Answer:

Not a solution

Step-by-step explanation:

In order for $x=-2$ to be a solution, the result after substituting must be 0:

$4(-2)^3-(-2)^2+16(-2)-4\stackrel{?}{=}0$

$4(-8)-4-32-4\stackrel{?}{=}0$

$-32-40\stackrel{?}{=}0$

$-72\neq0$

Therefore, $x=-2$ is not a solution to $4x^3-x^2+16x-4$

3 0

3 years ago