What is the correct answer to the math problem 20+30×0+1

Mathematics

2 answers:

Bumek [7]3 years ago

6 0

21
is what I got on my calculator

Alexeev081 [22]3 years ago

5 0

The answer is 21 because of PEMDAS
20+30x0+1
30x0=0 and you still have 20+1
and 20 +1 =21

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2. Round the following numbers to the nearest 10 thousand:

Vlad [161]

The values of the given numbers when it is rounded up to the nearest 10 thousands are:

990000
150000

<h3>What is rounding up in mathematics?</h3>

Rounding up can be described as the process that is been used in the mathematics which is been used in the estimation of a particular number in a context.

It should be noted that in rounding the a number up, it is required to look at the next digit at the right hand of the given figures in a case whereby the digit is less than 5,the digit can be rounded down, but in the case whereby the digit is more that 5 then it can be rounded up .

From the given values, we are given the 990,201 and 159,994 and if this were to rounded up to the nearest 10 thousand then we will start from the right hand sides and round down the values less than 5 and round up the values that is more that 5. and their values will be 990000

and 150000.

Read more about rounding up at:

brainly.com/question/28324571

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8 0

1 year ago

What is this Method???

Rina8888 [55]

Standard algorithm i think
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3 0

3 years ago

Read 2 more answers

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

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

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

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