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3 years ago

How to work the problem 2u²=3u+2

Mathematics

2 answers:

vlabodo [156]3 years ago

6 0

U= -1/2, 2

Hope it helps

solong [7]3 years ago

4 0

One way is to subtract 3u and 2 to the other side to make this equation quadratic. Set it equal to 0 and solve with the quadratic formula. You should get the answers Bubbies15 commented.

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Daniel had $25 to spend at the fair. If the admission to the fair is $4 and the rides cost $1.50 each, what is the greatest numb

mote1985 [20]

25= 1.50x+4
Subtract 4
21= 1.50x
Divide by 1.50
X=14 rides or less

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3 years ago

If f(x)=x^2 is horizontally compressed to g(x), which could be the equation of g(x)?

leva [86]

For the definition of horizontal compression, the function f(x) = x² is horizontally compressed to the function g(x) = (k · x)², for 0 < k < 1.

<h3>How to find the resulting equation after applying a compression</h3>

Here we must narrow a given function by a rigid operation known as compression. Rigid transformations are transformations in which Euclidean distances are conserved. In the case of functions, we define the horizontal compression in the following manner:

g(x) = f(k · x), for 0 < k < 1 (1)

If we know that f(x) = x², then the equation of g(x) is:

g(x) = (k · x)², 0 < k < 1

For the definition of horizontal compression, the function f(x) = x² is horizontally compressed to the function g(x) = (k · x)², for 0 < k < 1.

To learn more on rigid transformations: brainly.com/question/1761538

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2 years ago

Evaluate the following integral (Calculus 2) Please show step by step explanation!

barxatty [35]

Answer:

$\dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{x^2}{\sqrt{25-x^2}}\:\:\text{d}x$

Rewrite 25 as 5²:

$\implies \displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x$

Integration by substitution

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=5 \sin \theta$

$\begin{aligned}\implies \sqrt{5^2-x^2} & =\sqrt{5^2-(5 \sin \theta)^2}\\ & = \sqrt{25-25 \sin^2 \theta}\\ & = \sqrt{25(1-\sin^2 \theta)}\\ & = \sqrt{25 \cos^2 \theta}\\ & = 5 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=5 \cos \theta$

$\implies \text{d}x=5 \cos \theta\:\:\text{d}\theta$

Substitute everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x & = \int \dfrac{25 \sin^2 \theta}{5 \cos \theta}\:\:5 \cos \theta\:\:\text{d}\theta \\\\ & = \int 25 \sin^2 \theta\end{aligned}$

Take out the constant:

$\implies \displaystyle 25 \int \sin^2 \theta\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad \cos (2 \theta)=1 - 2 \sin^2 \theta$

$\implies \displaystyle 25 \int \dfrac{1}{2}(1-\cos 2 \theta)\:\:\text{d}\theta$

$\implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\cos kx$}\\\\$\displaystyle \int \cos kx\:\text{d}x=\dfrac{1}{k} \sin kx\:\:(+\text{C})$\end{minipage}}$

$\begin{aligned} \implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta & =\dfrac{25}{2}\left[\theta-\dfrac{1}{2} \sin 2\theta \right]\:+\text{C}\\\\ & = \dfrac{25}{2} \theta-\dfrac{25}{4}\sin 2\theta + \text{C}\end{aligned}$

$\textsf{Use the trigonometric identity}: \quad \sin (2 \theta)= 2 \sin \theta \cos \theta$

$\implies \dfrac{25}{2} \theta-\dfrac{25}{4}(2 \sin \theta \cos \theta) + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{25}{2}\sin \theta \cos \theta + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\sin \theta \cdot 5 \cos \theta + \text{C}$

$\textsf{Substitute back in } \sin \theta=\dfrac{x}{5} \textsf{ and }5 \cos \theta = \sqrt{25-x^2}:$

$\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\cdot \dfrac{x}{5} \cdot \sqrt{25-x^2} + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}$

$\textsf{Substitute back in } \theta=\arcsin \left(\dfrac{x}{5}\right) :$

$\implies \dfrac{25}{2} \arcsin \left(\dfrac{x}{5}\right) -\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}$

Take out the common factor 1/2:

$\implies \dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}$

Learn more about integration by trigonometric substitution here:

brainly.com/question/28157322

6 0

2 years ago