Solve this quadratic equation by completing the square.

Jeremy measured the room to be 10 feet long and 4.5 feet wide. He plans to purchase carpet for the entire room. The carpet cost

Andreyy89

Answer:

$205.73

Step-by-step explanation:

First we find the square area,

multiply 10 by 4.5 and we get 45

we convert feet to meters by dividing it by 3.281

45/3.281=13.71533

we the multiply 13.71533 by 15 to get the total price

we get 205.72996 and round it to 205.72

5 0

3 years ago

14. Team 8A is selling snacks at the football game to raise money for a field trip. They are selling popcorn for $2.50 and pretz

Mamont248 [21]

Answer:

40 popcorns and 90 pretzels sold.

============

Let the number of popcorns is x and pretzels is y.

Set up equations:

x + y = 130
2.5x + 2y = 280

Solve by elimination, multiply the first equation by 2 and subtract from the second equation:

2.5x + 2y - 2x - 2y = 280 - 2*130
0.5x = 280 - 260
0.5x = 20
x = 20/0.5
x = 40

Find y:

40 + y = 130
y = 130 - 40
y = 90

5 0

2 years ago

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Linear or nonlinear!?

NeTakaya

Answer:

nonlinear

Step-by-step explanation:

7 0

3 years ago

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Area of the bounded curves y=x^2, y=√(7+x)

N76 [4]

Answer:

$\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = 5.74773$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

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

Integration

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]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

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

Step-by-step explanation:

Step 1: Define

$\displaystyle \left \{ {{y = x^2} \atop {y = \sqrt{7 + x}}} \right.$

Step 2: Identify

Graph the systems of equations - see attachment.

Top Function: $\displaystyle y = \sqrt{7 + x}$

Bottom Function: $\displaystyle y = x^2$

Bounds of Integration: [-1.529, 1.718]

Step 3: Integrate Pt. 1

Substitute in variables [Area of a Region Formula]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{1.718}_{-1.529} {\sqrt{7 + x}} \, dx - \int\limits^{1.718}_{-1.529} {x^2} \, dx$
[Right Integral] Integration Rule [Reverse Power Rule]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{1.718}_{-1.529} {\sqrt{7 + x}} \, dx - \frac{x^3}{3} \bigg| \limits^{1.718}_{-1.529}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{1.718}_{-1.529} {\sqrt{7 + x}} \, dx - 2.88176$

Step 4: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 7 + x$
[u] Basic Power Rule [Derivative Rule - Addition/Subtraction]: $\displaystyle du = dx$
[Limits] Switch: $\displaystyle \left \{ {{x = 1.718 ,\ u = 7 + 1.718 = 8.718} \atop {x = -1.529 ,\ u = 7 - 1.529 = 5.471}} \right.$

Step 5: Integrate Pt. 3

[Integral] U-Substitution: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{8.718}_{5.471} {\sqrt{u}} \, du - 2.88176$
[Integral] Integration Rule [Reverse Power Rule]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = \frac{2x^\Big{\frac{3}{2}}}{3} \bigg| \limits^{8.718}_{5.471} - 2.88176$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = 8.62949 - 2.88176$
Simplify: $\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = 5.74773$

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

Unit: Integration

5 0

3 years ago

If the average daily high temperature for the week was 28C, find the missing temperature from the following list: 25C, 26C, 31C,

DiKsa [7]

I believe the answer would be 27C.

5 0

3 years ago

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