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3 years ago

What percent of 300 is 90. Help please

2 answers:

3 0

30%.

Well, the simplest way to do it is to put it as a fraction.
90/300.

This is the same as 3/10, which is the same as .30, which is the same as 30% :P

Hope this helped!

zhuklara [117]3 years ago

3 0

$\frac{90}{300}\cdot100\%=\frac{3}{10}\cdot100\%=3\cdot10\%=30\%$

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The area of the triangle is eqaula to the sqare show that x²-3x-2=0

agasfer [191]

Answer:

Step-by-step explanation:

Area of square = area of triangle

$side*side = \dfrac{1}{2}b*h\\\\\\x*x=\dfrac{1}{2}*(x+1)(x+2)\\x^{2}=\dfrac{1}{2}[x*x +x*2+1*x+1*2]\\\\\\x^{2}=\dfrac{1}{2}[x^{2}+2x+x+2]\\\\\\x^{2}=\dfrac{1}{2}[x^{2}+3x+2]\\\\\\Multiply \ both \ sides \ by \ 2\\\\2x^{2}=x^{2}+3x+2\\\\2x^{2}-x^{2}-3x-2 = 0$

x² - 3x - 2 = 0

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2 years ago

Please show your workings

GarryVolchara [31]

13% of 4,760,000 = 618,800
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Hope it helps!

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3 years ago

x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco

igomit [66]

Answer:

$x=-cos(t)+2sin(t)$

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

$a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0$

Will have a characteristic equation of the form:

$a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0$

Where solutions $r_1,r_2...,r_n$ are the roots from which the general solution can be found.

For real roots the solution is given by:

$y(t)=c_1e^{r_1t} +c_2e^{r_2t}$

For real repeated roots the solution is given by:

$y(t)=c_1e^{rt} +c_2te^{rt}$

For complex roots the solution is given by:

$y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)$

Where:

$r_1_,_2=\lambda \pm \mu i$

Let's find the solution for $x''+x=0$ using the previous information:

The characteristic equation is:

$r^{2} +1=0$

So, the roots are given by:

$r_1_,_2=0\pm \sqrt{-1} =\pm i$

Therefore, the solution is:

$x(t)=c_1cos(t)+c_2sin(t)$

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of $x(t)$ in order to find the constants $c_1$ and $c_2$ :

$x'(t)=-c_1sin(t)+c_2cos(t)$

Evaluating the initial conditions:

$x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1$

And

$x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2$

Now we have found the value of the constants, the solution of the second-order IVP is:

$x=-cos(t)+2sin(t)$

3 0

3 years ago

A set of ordered pairs is called ordered pairs is called

Amiraneli [1.4K]

A set of ordered pairs is called a relation. The set of all first components of the ordered pairs of a relation is the domain of the relation, and the set of all second components of the ordered pairs is the range of the relation. :)

4 0

3 years ago

Aadya and Ethan like playing with remote-control cars. There's a straight track at the

e-lub [12.9K]

The distance between the two cars is now 320 yards.

<h3 /><h3><u>Distances</u></h3>

To determine what is the distance between the two cars now if they did not collide the following calculation must be performed:

500 - (500 - 405) - (500 - 415) = X
500 - 95 - 85 = X
500 - 180 = X
320 = X

Therefore, the distance between the two cars is now 320 yards.

Learn more about distances in brainly.com/question/989117

8 0

3 years ago