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3 years ago

How does the percent equation relate proportional quantities?

Mathematics

1 answer:

Julli [10]3 years ago

5 0

Answer:

Use cross product to determine if the two ratios form a proportion. Here we can see that 2/16 and 5/40 are proportions since their cross products are equal. Percent means hundredths or per hundred and is written with the symbol, %. Percent is a ratio were we compare numbers to 100 which means that 1% is 1/100.

Step-by-step explanation:

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−9x (5x+4 ) +10 (x+3) can someone please solve his for me thank u step by step please i really need to find out how to do it

Ivahew [28]

Answer:

-45 $x^{2}$ -26x+30

Step-by-step explanation:

So first you let's divide this equation into three parts.

Part one -9x(5x+4)

Step one: you have to distribute the -9x to the numbers in the parentheses. That would leave us with a -9x*5x = -45 $x^{2}$ and -9x*4 = -36x
Step two: Put the answers together. That would leave us with -45 $x^{2}$ -36x.

Part two 10(x+3)

Step one: you have to distribute the 10 to the numbers in the parentheses. That would leave us with a 10*x = 10x and 10*3 = 30
Step two: Put the answers together. That would leave us with 10x+30

Part three solve

Step one: combine and put together what you have solved for. That would leave us with a -45 $x^{2}$ -36x+ 10x+30
Step two: combine like terms. The like terms here are -36x and 10x. When you combine them you would get -26x.
Step three: Write the equation in standard form. Therefore the answer is -45 $x^{2}$ -26x+30.

Let me know if anything is confusing!

5 0

3 years ago

If F(x) = log 4 3x, find F -1(x).

jenyasd209 [6]

Answer:

h(-4) = -4

Step-by-step explanation:

Since x is given (x = -4). Then, all you need to do is log in -4 for each value of x

h(-4) =

−

(

−

)

-3(-4)

h(-4) = -(16) + 12 = -4

4 0

3 years ago

Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck

Nimfa-mama [501]

$\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}$
$\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}$

Now, isolate the variables, so you can integrate.
$(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}$
$\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}$
$2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C$

$4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C$
$y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C$
$(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C$

$y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$
$y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}$

At x = 1, y = 0.
$0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$
$-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0$
$\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}$

$4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C$
$0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C$
$C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}$
$C = -\frac{\pi}{2}$

$\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}$