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4 years ago

Which equation represents the equation of the parabola with focus (-3 3) and directrix y=7?

Mathematics

1 answer:

Artemon [7]4 years ago

6 0

Answer:

The equation $y=\frac{-x^2-6x+31}{8}$ represents the equation of the parabola with focus (-3, 3) and directrix y = 7.

Step-by-step explanation:

To find the equation of the parabola with focus (-3, 3) and directrix y = 7. We start by assuming a general point on the parabola (x, y).

Using the distance formula $d = \sqrt {\left( {x_1 - x_2 } \right)^2 + \left( {y_1 - y_2 } \right)^2 }$ , we find that the distance between (x, y) is

$\sqrt{(x+3)^2+(y-3)^2}$

and the distance between (x, y) and the directrix y = 7 is

$\sqrt{(y-7)^2}$ .

On the parabola, these distances are equal so, we solve for y:

$\sqrt{(x+3)^2+(y-3)^2}=\sqrt{(y-7)^2}\\\\\left(\sqrt{\left(x+3\right)^2+\left(y-3\right)^2}\right)^2=\left(\sqrt{\left(y-7\right)^2}\right)^2\\\\x^2+6x+y^2+18-6y=\left(y-7\right)^2\\\\x^2+6x+y^2+18-6y=y^2-14y+49\\\\y=\frac{-x^2-6x+31}{8}$

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Answer:

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4 years ago

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3 years ago

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Integrating sums of functions

Andrei [34K]

Answer:

(a) -12

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Calculus

Integrals

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

Integration Property [Swapping Limits]: $\displaystyle \int\limits^b_a {f(x)} \, dx = -\int\limits^a_b {f(x)} \, dx$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

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

Integration Property [Splitting Integral]: $\displaystyle \int\limits^c_a {f(x)} \, dx = \int\limits^b_a {f(x)} \, dx + \int\limits^c_b {f(x)} \, dx$

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

Step-by-step explanation:

Step 1: Define

$\displaystyle \int\limits^6_4 {f(x)} \, dx = 5$

$\displaystyle \int\limits^4_{10} {f(x)} \, dx = 8$

$\displaystyle \int\limits^{10}_6 {[4f(x) + 10]} \, dx$

Step 2: Solve Pt. 1

[Integral] Rewrite [Integration Property - Addition]: $\displaystyle \int\limits^{10}_6 {[4f(x) + 10]} \, dx = \int\limits^{10}_6 {4f(x)} \, dx + \int\limits^{10}_6 {10} \, dx$
[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^{10}_6 {[4f(x) + 10]} \, dx = 4\int\limits^{10}_6 {f(x)} \, dx + 10\int\limits^{10}_6 {} \, dx$

Step 3: Redefine

Manipulate the given integral values.

[Integrals] Combine [Integration Property - Splitting Integral]: $\displaystyle \int\limits^6_4 {f(x)} \, dx + \int\limits^4_{10} {f(x)} \, dx = \int\limits^6_{10} {f(x)} \, dx$
[Integral] Rewrite: $\displaystyle \int\limits^6_{10} {f(x)} \, dx = \int\limits^6_4 {f(x)} \, dx + \int\limits^4_{10} {f(x)} \, dx$
[Integral] Substitute in integrals: $\displaystyle \int\limits^6_{10} {f(x)} \, dx = 5 + 8$
[Integral] Add: $\displaystyle \int\limits^6_{10} {f(x)} \, dx = 13$
[Integral] Rewrite [Integration Property - Swapping Limits]: $\displaystyle -\int\limits^{10}_6 {f(x)} \, dx = 13$
[Integral] [Division Property of Equality] Isolate integral: $\displaystyle \int\limits^{10}_6 {f(x)} \, dx = -13$

Step 4: Solve Pt. 2

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[Integral] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle \int\limits^{10}_6 {[4f(x) + 10]} \, dx = 4(-13) + 10(x) \bigg| \limits^{10}_6$
[Integral] Evaluate [Integration Rule - FTC 1]: $\displaystyle \int\limits^{10}_6 {[4f(x) + 10]} \, dx = 4(-13) + 10(10 - 6)$
[Integral] (Parenthesis) Subtract: $\displaystyle \int\limits^{10}_6 {[4f(x) + 10]} \, dx = 4(-13) + 10(4)$
[Integral] Multiply: $\displaystyle \int\limits^{10}_6 {[4f(x) + 10]} \, dx = -52 + 40$
[Integral] Add: $\displaystyle \int\limits^{10}_6 {[4f(x) + 10]} \, dx = -12$

Topic: AP Calculus AB/BC

Unit: Integration

Book: College Calculus 10e

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3 years ago

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Answer:

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To round to four significant figures, look at the fifth significant figure. It's a 1 , so round down.

To round to two significant figures, look at the third significant figure. It's an 8 , so round up.

Step-by-step explanation:

use this trick

bye

mark brainliest

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