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2 years ago

Hi! Can you join my quizizz? The code is 5373177. The website is quizizz.com

Mathematics

2 answers:

Xelga [282]2 years ago

6 0

Answer:

ooooooooooooh maybe

Step-by-step explanation:

Cool

rosijanka [135]2 years ago

6 0

plsssssssss and thxxxxxxx

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Find the general solution of the given differential equation. cos^2(x)sin(x)dy/dx+(cos^3(x))y=1 g

eimsori [14]

If the given differential equation is

$\cos^2(x) \sin(x) \dfrac{dy}{dx} + \cos^3(x) y = 1$

then multiply both sides by $\frac1{\cos^2(x)}$ :

$\sin(x) \dfrac{dy}{dx} + \cos(x) y = \sec^2(x)$

The left side is the derivative of a product,

$\dfrac{d}{dx}\left[\sin(x)y\right] = \sec^2(x)$

Integrate both sides with respect to $x$ , recalling that $\frac{d}{dx}\tan(x) = \sec^2(x)$ :

$\displaystyle \int \frac{d}{dx}\left[\sin(x)y\right] \, dx = \int \sec^2(x) \, dx$

$\sin(x) y = \tan(x) + C$

Solve for $y$ :

$\boxed{y = \sec(x) + C \csc(x)}which follows from [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}$ .

7 0

2 years ago

Which two theorems would justify that m∠4 = m∠6, given that m∠5 = m∠6 in the diagram below?

My name is Ann [436]

Answer: D) vertical angles theorem, alternate interior angles theorem

Angle 5 = Angle 6 by the alternate interior angles theorem

Angle 5 = angle 4 by the vertical angles theorem

By the transitive property, we can then say angle 4 = angle 6. These angles are also corresponding angles.

We won't use the angle addition theorem or the right angles theorem.

4 0

3 years ago

Read 2 more answers

24 points scored in 4 games; 48 points scored in 10 games... are they equivalent?

ch4aika [34]

Hello.

24 points scored in 4 games = 24 * 4.

24 * 4 = 96

48 points scored in 10 games = 48 * 10

48 * 10 = 480

<em>96≠480, they are not equivalent.</em>

5 0

3 years ago

I don't get these what Do I do

max2010maxim [7]

Answer:

No idea

Step-by-step explanation:

3 0

2 years ago

How do you find a vector that is orthogonal to 5i + 12j ?

Rashid [163]

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}$

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

4 0

3 years ago