If the masses of objects increase and their distance from each other remains the same, then the force of gravity between the two

To find the highest point, we want to find the vertex of the parabola. Since the function will be negative, the vertex will represent the highest point.

To find the vertex we want this equation in vertex form:

y=a(x−h)2+k

and the vertex will be (h,k)

We can expand the equation h(x)=-(x+1)(x-7) to be:

$h(x) = -x^{2} +6x+7$

To get the vertex form we need to complete the square.

The vertex form is :

$-(x-3)^{2} +16$

so our vertex is (3,16)

This means the highest point will be 16 when x = 3

6 0

3 years ago

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The x - coordinate of B' is _____ . The y - coordinate of B' is _____ please help ;-;

Ilya [14]

Answer:

B' (2, 5 )

Step-by-step explanation:

Under a counterclockwise rotation about the origin of 270°

a point (x, y ) → (y, - x ) , then

B(- 5, 2 ) → B' (2, 5 )

7 0

3 years ago

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