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3 years ago

10

A dialonal cross section of a sphere produces which two dimesional shape?

1 answer:

stellarik [79]3 years ago

6 0

Regardless of how the sphere<span> is cut, it will form a circle</span>

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27 is 3 as many times aswhat number

Andru [333]

27 is 3 as many times the number 9

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3 years ago

Read 2 more answers

Help evaluating the indefinite integral

Dafna11 [192]

Answer:

$\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \boxed{ -\sqrt{4 - x^2} + C }$

General Formulas and Concepts:
<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]:
$\displaystyle (cu)' = cu'$

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$
Derivative Rule [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 Property [Multiplied Constant]:
$\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Methods: U-Substitution and U-Solve

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify given.</em>

<em /> $\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution/u-solve</em>.

Set <em>u</em>:
$\displaystyle u = 4 - x^2$
[<em>u</em>] Differentiate [Derivative Rules and Properties]:
$\displaystyle du = -2x \ dx$
[<em>du</em>] Rewrite [U-Solve]:
$\displaystyle dx = \frac{-1}{2x} \ du$

<u>Step 3: Integrate Pt. 2</u>

[Integral] Apply U-Solve:
$\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \int {\frac{-x}{2x\sqrt{u}}} \, du$
[Integrand] Simplify:
$\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \int {\frac{-1}{2\sqrt{u}}} \, du$
[Integral] Rewrite [Integration Property - Multiplied Constant]:
$\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \frac{-1}{2} \int {\frac{1}{\sqrt{u}}} \, du$
[Integral] Apply Integration Rule [Reverse Power Rule]:
$\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = -\sqrt{u} + C$
[<em>u</em>] Back-substitute:
$\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \boxed{ -\sqrt{4 - x^2} + C }$

∴ we have used u-solve (u-substitution) to <em>find</em> the indefinite integral.

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Learn more about integration: brainly.com/question/27746495

Learn more about Calculus: brainly.com/question/27746485

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

Unit: Integration

5 0

2 years ago

A package of 4 pairs of insulated gloves cost 35.96

tino4ka555 [31]

Answer:

You need to add more to this question but I am assuming you want to know how much each pair of gloves is?

Step-by-step explanation:

Divide 35.96 by 4

35.96÷4= $8.99 per pair of insulated gloves

6 0

3 years ago

Math 8: Linear Functions, Part 2

Studentka2010 [4]

Answer:

$\boxed{\text{1. y + 5 = -4(x - 3); \qquad 2. y - 8 = x + 1}}$

Step-by-step explanation:

Question 1

The point-slope formula for a straight line is

y – y₁ = m(x – x₁)

x₁ = 3; y₁ = -5; m = -4

Substitute the values

$\boxed{\textbf{y + 5 = -4(x - 3)}}$

The diagram shows the graph of equation 1 (red) with slope -4 passing through (3,-5).

Question 2

x₁ = -1; y₁ = 8; m = 1

Substitute the values

$\boxed{\textbf{y - 8 = x + 1}}$

The diagram shows the graph of equation 2 (green) with slope 1 passing through (-1,8).

3 0

3 years ago

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

<em>A.</em>

<em>The student made an error in step 3 because a is positive in Quadrant IV; therefore, </em>

<em /> $cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

<em>The student's mistake is that a is positive in quadrant iv and his error is in step 3</em>

3 0

3 years ago