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marshall27 [118]

3 years ago

15

Andrea does not understand why if x is a solution to the equation 3-x=4 then x is also a solution to the equation 12=9-3x. Write

a convincing explanation as to why this is true

1 answer:

Strike441 [17]3 years ago

6 0

Answer:

Both equations are same

Step-by-step explanation:

The reason why this is true is not far fetched

looking at the first equation, we have;

3-x = 4

we can also factorize the second equation to look this way

By this, we shall be having

3(4) = 3(3-x)

By the time we take out the 3 from both sides, we will have the initial equation

So what we are saying is that both equations are the same

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To stay properly hydrated, a person should drink 32 ounces of water for every 60 minutes of exercise. How much water should Damo

EleoNora [17]

72 as Damon spend 2 and 15 minutes exercising so for the two dull hours you add 64 now that done and as 15 is 1/4 of 60 you add 8 which is 1/4 of 32 and get 72

6 0

3 years ago

Find a number C such that the polynomial p(x)= -x+4x^2+Cx^3-8x^4

Ksenya-84 [330]

Answer:

C = 2

Step-by-step explanation:

$p(x)= -x+4x^2+Cx^3-8x^4 \\ \\ \because \: given \: polynmial \: has \: zero \: at \: \\x = \frac{1}{4} \\ \\ \implies \: p\bigg(\frac{1}{4} \bigg) = 0...(1) \\ \\ plug \: x = \frac{1}{4} \: in \: p(x) \: we \: find \\ \\ p \bigg(\frac{1}{4} \bigg) = - \frac{1}{4} + 4 {\bigg(\frac{1}{4} \bigg)}^{2} + c {\bigg(\frac{1}{4} \bigg)}^{3} - 8 {\bigg(\frac{1}{4} \bigg)}^{4} \\ \\ 0 = - \frac{1}{4} + \cancel 4 \times {\frac{1}{\cancel{16} } } + c {\bigg(\frac{1}{64} \bigg)} - \cancel 8 \times \frac{1}{\cancel{256} } \\ \\0 =\cancel{ - \frac{1}{4}} + \cancel {{\frac{1}{4} }} + c {\bigg(\frac{1}{64} \bigg)} - \times \frac{1}{32} \\ \\0 = c {\bigg(\frac{1}{64} \bigg)} - \frac{1}{32} \\ \\c {\bigg(\frac{1}{64} \bigg)} = \frac{1}{32} \\ \\ c = \frac{1}{32} \times 64 \\ \\ c = 2$

5 0

3 years ago

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

Y=-1/3x^2-4x-5 in vertex form

grigory [225]

Answer:

$Y=-\frac{1}{3}(x+6)^2+7$ [Vertex form]

Step-by-step explanation:

Given function:

$Y=-\frac{1}{3}x^2-4x-5$

We need to find the vertex form which is.,

$y=a(x-h)^2+k$

where $(h,k)$ represents the co-ordinates of vertex.

We apply completing square method to do so.

We have

$Y=-\frac{1}{3}x^2-4x-5$

First of all we make sure that the leading co-efficient is =1.

In order to make the leading co-efficient is =1, we multiply each term with -3.

$-3\times Y=-3\times\frac{1}{3}x^2-(-3)\times4x-(-3)\times 5$

$-3Y=x^2+12x+15$

Isolating $x^2$ and $x$ terms on one side.

Subtracting both sides by 15.

$-3Y-15=x^2+12x-15-15$

$-3Y-15=x^2+12x$

In order to make the right side a perfect square trinomial, we will take half of the co-efficient of $x$ term, square it and add it both sides side.

square of half of the co-efficient of $x$ term = $(\frac{1}{2}\times 12)^2=(6)^2=36$

Adding 36 to both sides.

$-3Y-15+36=x^2+12x+36$

$-3Y+21=x^2+12x+36$

Since $x^2+12x+36$ is a perfect square of $(x+6)$ , so, we can write as:

$-3Y+21=(x+6)^2$

Subtracting 21 to both sides:

$-3Y+21-21=(x+6)^2-21$

$-3Y=(x+6)^2-21$

Dividing both sides by -3.

$\frac{-3Y}{-3}=\frac{(x+6)^2}{-3}-\frac{21}{-3}$

$Y=-\frac{1}{3}(x+6)^2+7$ [Vertex form]

8 0

3 years ago

write an equation in slope-intercept form for a line that goes through (9,-2) and is parallel to y=7x+2

lozanna [386]

Answer:

y=7x-65

Step-by-step explanation:

3 0

2 years ago