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2 years ago

I need help on this...how do i do it

Mathematics

2 answers:

Ugo [173]2 years ago

6 0

Answer:

Step-by-step explanation:

You can first reduce this fraction by dividing both the numerator and denominator by the Greatest Common Factor of 549 and 999. I know that

61/111

is the same as

61÷111

Then using

Long Division for 61 divided by 111

and rounding to a Max of 4 Decimal Places gives me

=0.5495/.5495

<h3>then move the decimal point a digit to the <u><em>RIGHT!!!!</em></u></h3>

svetlana [45]2 years ago

3 0

The answer is A I think hope this helps

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Which algebraic expression is a polynomial? 3m2n-2m/n+1/n

Lesechka [4]

Answer:

d) $7mn+\frac{3m}{2}+\frac{5n}{4}$

Step-by-step explanation:

Which algebraic expression is a polynomial?

$a)3m^2n-\frac{2m}{n}+\frac{1}{n} \\b)\frac{2mn}{5}-\frac{\sqrt{m} }{4} +4m^5\\c)\frac{4m^3}{n^2}-3mn^5+\sqrt{8}\\d)7mn+\frac{3m}{2}+\frac{5n}{4}$

Answer: A polynomial is an expression with variables and coefficients separated by addition and subtraction. The following are used when determining if an expression is a polynomial:

i) Variables of a polynomial must not have negative exponent.

ii) Variables of a polynomial must not have fractional exponent.

iii) Dividing by an exponent is not allowed

iv) Exponent radicals are not allowed.

a) $3m^2n-\frac{2m}{n}+\frac{1}{n}$ is not a polynomial because dividing by an exponent is not allowed.

b) $\frac{2mn}{5}-\frac{\sqrt{m} }{4} +4m^5$ , is not a polynomial because exponent radicals are not allowed (√m)

c) $\frac{4m^3}{n^2}-3mn^5+\sqrt{8}$ is not a polynomial because dividing by an exponent is not allowed.

d) $7mn+\frac{3m}{2}+\frac{5n}{4}$ is a polynomial

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3 years ago

The test scores of students in a math class are 56, 42, 50, 63, 58, 41, 50, 48, 68, and 48. What is the mean of this set of data

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To find the mean, add up all the test scores and divide by the number of tests:
524/10 = 52.4.

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4 years ago

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Show that cos(3x) = cos3(x) − 3 sin2(x)cos(x). (Hint: Use cos(2x) = cos2(x) − sin2(x) and the cosine sum

otez555 [7]

<u>Step-by-step explanation:</u>

To prove:

$\cos 3x=\cos^3x-3\sin^2 x\cos x$

Identities used:

$\cos(A+B)=\cos A\cos B+\sin A\sin B$ ......(1)

$\sin 2A=2\sin A\cos A$ ........(2)

$\cos 2A=\cos^2A-\sin^2 A$ .......(3)

Taking the LHS:

$\Rightarrow \cos 3x=\cos (x+2x)$

Using identity 1:

$\Rightarrow \cos (x+2x)=\cos x\cos 2x-\sin x\sin 2x$

Using identities 2 and 3:

$\Rightarrow \cos x(\cos ^2x-\sin^2 x)-\sin x(2\sin x\cos x)\\\\\Rightarrow \cos^3x-\sin^2x\cos x-2\sin^2 x\cos x\\\\\Rightarrow \cos^3x-3\sin^2x\cos x$

As, LHS = RHS

Hence proved

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3 years ago