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3 years ago

RRC #2 What is the radius of this circle? The radius of the circle is 24cm

Mathematics

1 answer:

poizon [28]3 years ago

7 0

Answer:

25cm

Step-by-step explanation:

The radius of circles is 7cm,24cm

Area of smaller circle is πr

⟹π×7

⟹49π

Area of Larger circle is πr

⟹π×24

⟹576π

Area of required circle ⟹576π+49π

⟹625π

Radius of circle is given by

625

=25

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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 single gram of a certain metallic substance has 0.52 g of copper and 0.25 g of zinc the remaining portion of the substance is

andrezito [222]

Ben's estimate gives 7 g of nickel; the actual amount is 8.03 g.

In 1 g of the substance, there is 0.52 g of copper and 0.25 g of zinc; this gives
0.52+0.25 = 0.77 g of the substance.

The remaining part of the substance is nickel:
1-0.77 = 0.23 g of nickel.

Using Ben's estimate, 0.2 g of nickel per gram of substance, we have
0.2(35) = 7 g of nickel in 35 g of the substance.

The actual amount is 0.23(35) = 8.03 g of nickel in 35 g of the substance.

6 0

3 years ago

How do I find the area for this?

Olegator [25]

Answer: when u doing area know you are multiplying hope i help

Step-by-step explanation:

3 0

3 years ago

You work 4 hours on Monday, 4.5 hours on Tuesday, 7.25 hours on Thursday, and 12 hours on Saturday. If you get paid $11.52 per h

likoan [24]

Answer:

319.68

Step-by-step explanation:

you multyply 11.52 per each hour and then add everything up

Wish this helped you

7 0

3 years ago

Solve 33 1/2 of 48 Plz thanks <3

SIZIF [17.4K]

Answer:

1608

Step-by-step explanation:

5 0

3 years ago

RRC #2 What is the radius of this circle? The radius of the circle is 24cm​

RRC #2 What is the radius of this circle? The radius of the circle is 24cm