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3 years ago

THE IMAGE DOWN BELOW MATH PLEASE HELP ME WITH THIS ALEGEBRA IM CRYING YALL

Mathematics

1 answer:

Fittoniya [83]3 years ago

4 0

Answer:

25% increase

Step-by-step explanation:

final value (20) -minus- starting value (16)

-------------------divided by ---------------- times * 100

starting value (20)

--- x 100

25%

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Find the error & find the correct answer 2In(x)=In(3x)-[In(9)-2In(3)] In(x^2)=In(3x)-[In(9)-In(9)] In(x^2)=In(3x)-0 In(x^2)=

Art [367]

Answer:

Error: $lnx^2=ln 3x$ not $ln\frac{3x}{0}$

Solution:x=0 and 3

Step-by-step explanation:

We have to find the error and correct answer

Given: $2ln x=ln(3x)-[ln9-2ln(3)]$

$lnx^2=ln(3x)-[ln9-ln3^2]$

Using the formula

$alog b=logb^a$

$lnx^2=ln(3x)-[ln9-ln9]$

$lnx^2=ln(3x)-0$

$lnx^2=ln(3x)$

$x^2=3x$

$x^2-3x=0$

$x(x-3)=0$

Therefore, x=0 and x=3

But last step in the given solution

$lnx^2=ln\frac{3x}{0}=\infty$

It is wrong this property is used when

$log m-log n$ then

$log\frac{m}{n}$

Hence, the student wrote $lnx^2=ln\frac{3x}{0}$ instead of $lnx^2=ln3x$ and solution is given by

x=0 and x=3

4 0

3 years ago

The diameter of a circle is 18 centimeters. what is the radius? d=18 cm give the exact answer in simplest form.

goldenfox [79]

The diameter of the circle is 18 cm.

The radius is one half the diameter.
Therefore the radius is 18/2 = 9 cm

Answer: 9 cm

4 0

3 years ago

Please can anyone help?

Elena L [17]

Answer:

9/5x^3

Step-by-step explanation:

i hope that is the answer

7 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

The _____ of two sets X and Y is denoted by X-Y

lidiya [134]

Answer:

Cartesian product

Step-by-step explanation:

The Cartesian product of two sets, X and Y, denoted by X × Y, is the set of all ordered pairs (x, y), where x is an element of X and y is an element of Y: 8 (2.4.1) X × Y = { (x, y) ∣ x ∈ X ∧ y ∈ Y } For example, if Children = { Peter, Mark, Mary }, and Parents = { Paul, Jane, Mark, Mary }, then

5 0

2 years ago