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3 years ago

If the scale factor between two circles is 7a/9b what is the ratio of their areas?

Mathematics

2 answers:

RideAnS [48]3 years ago

5 0

. 49a2/81b2
I think the answer is B)

Dvinal [7]3 years ago

5 0

Answer: The correct option is (B) $\dfrac{49a^2}{81b^2}.$

Step-by-step explanation: Given that the scale factor between two circles is $\dfrac{7a}{9b}$ .

We are to find the ratio of the areas of the two circles.

Let r and R be the radii of the two circles and A and B be the corresponding areas.

We know that the scale factor is the ratio of the radii of the two circles.

So, we have

$\dfrac{r}{R}=\dfrac{7a}{9b}\\\\\\\Rightarrow \dfrac{r^2}{R^2}=\dfrac{49a^2}{81b^2}\\\\\\\Rightarrow \dfrac{\pi r^2}{\pi R^2}=\dfrac{49a^2}{81b^2}\\\\\\\Rightarrow \dfrac{A}{B}=\dfrac{49a^2}{81b^2}.$

Thus, the required ratio of the areas of the two circles is $\dfrac{49a^2}{81b^2}.$

Option (B) is CORRECT.

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If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

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

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

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

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

Unit: Integration

4 0

3 years ago

A 60 B 30 C 1 20 D 15 fastest answer gets brainiest

Anuta_ua [19.1K]

Answer:

i think it would be A

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

1.9 as a mixed number in simplest form

Yuri [45]

1.9 = 1 9/10 in simplest form

5 0

3 years ago

i am a number less than 3,000. when you divide me by 32,my remainder is 30. when you divide me by 58,my remainder is 44. what nu

Marat540 [252]

32(24)+30=58(13)+44=798

I am 798

3 0

3 years ago

Can someone tell me if i got this right? its all i need

Alexxx [7]

Answer:

No, the answer is 120 plants per square foot

Step-by-step explanation:

What I did was first figure up the area of the triangle. You want to do this because the area will tell you how many square feet are in the garden. In order to find this, you need to use this formula: 1/2 b·h

Plug in the numbers like this: 1/2·15·8

15 is the base of the triangle, while 8 is the height of the triangle. Now solve your equation.

15·8=120

1/2·120=60

Area = 60

Now you know that the area is 60ft². So there are 60 square feet in the whole garden. But, we're not done yet. Jessica wants to put 2 plants in every 1 square foot. So now, we need to multiply 60 by 2 in order to get 2 plants in every 1 square foot.

60·2=120

Jessica can put 120 plants in the garden if she is wanting 2 plants per square foot.

I hope this makes sense and is easy to understand. Please let me know if you need any more help or clarification. I'm always happy to help! Have a great day and good luck!!

4 0

3 years ago