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vlada-n [284]
3 years ago
9

NEED HELP WORTH 100 POINTS SHOW YOUR WORK PLEASE

Mathematics
1 answer:
tia_tia [17]3 years ago
5 0

Answer:

Q1:

-5\sqrt{27c^2}

-5\sqrt{27}\sqrt{c^2}

-5\times\sqrt{9} \sqrt{3} \sqrt{c^2}

-5\times3\sqrt{3}\sqrt{c^2}

-5\times3\sqrt{3}c

-15\sqrt{3}c

Q2:

5\sqrt{72}-3\sqrt{32}

5\sqrt{36} \sqrt{2} -3\sqrt{16} \sqrt{2}

5\times6\sqrt{2}-3\times4\sqrt{2}

30\sqrt{2}-12\sqrt{2}

18\sqrt{2}

Q3:

\sqrt{108yz}+3\sqrt{98yz}+2\sqrt{75yz}

\sqrt{36} \sqrt{3} \sqrt{yz} +3\sqrt{49} \sqrt{2} \sqrt{yz} +2\sqrt{25} \sqrt{3} \sqrt{yz}

6\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}+10\sqrt{3}\sqrt{yz}

16\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}

Q4:

\sqrt{15n^2}\sqrt{10n^3}

\sqrt{15}n\sqrt{10n^3}

\sqrt{15}n\sqrt{10}\sqrt{n^2}\sqrt{n}

\sqrt{15}\sqrt{10}n^2\sqrt{n}

\sqrt{5}\sqrt{3}\sqrt{5}\sqrt{2}n^2\sqrt{n}

5\sqrt{3}\sqrt{2}n^2\sqrt{n}

5\sqrt{6}n^2\sqrt{n}

Q5:

\frac{\sqrt{3x^2y^3}}{4\sqrt{5xy^3}}

\frac{\sqrt{3}xy\sqrt{y}}{4\sqrt{5}y\sqrt{x}\sqrt{y}}

Cancel off \sqrt{x}:

\frac{\sqrt{3}y\sqrt{x}\sqrt{y}}{4\sqrt{5}y\sqrt{y}}

Cancel off y:

\frac{\sqrt{3}\sqrt{x}\sqrt{y}}{4\sqrt{5}\sqrt{y}}

Cancel off \sqrt{y} :

\frac{\sqrt{3}\sqrt{x}}{4\sqrt{5}}

Multiply \sqrt{5} on the numerator and denominator:

\frac{\sqrt{3}\sqrt{x}\sqrt{5}}{4\sqrt{5}\sqrt{5}}

\frac{\sqrt{15}\sqrt{x}}{20}

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Answer:

5 km

Step-by-step explanation:

the scans means that any distance on the map is 50000 times larger in real life.

so, 10 cm on the map is

10×50000 = 500000 cm in real life.

now, we know that 100cm = 1 m

so, it is 500000/100 = 5000 m in real life.

and there are 1000 m in 1 km.

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7 0
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Determine whether each of the binary relation R defined on the given sets A is reflexive, symmetric, antisymmetric, or transitiv
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Answer:

In explanation

Please let me know if something doesn't make sense.

Step-by-step explanation:

a)

*This relation is not reflexive.

0 is an integer and (0,0) is not in the relation because 0(0)>0 is not true.

*This relation is symmetric because if a(b)>0 then b(a)>0 since multiplication is commutative.

*This relation is transitive.

Assume a(b)>0 and b(c)>0.

Note: This means not a,b, or c can be zero.

Therefore we have abbc>0.

Since b^2 is positive then ac is positive.

Since a(c)>0, then (a,c) is in R provided (a,b) and (b,c) is in R.

*The relation is not antisymmretric.

(3,2) and (2,3) are in R but 3 doesn't equal 2.

b)

*This relation is reflective.

Since a^2=a^2 for any a, then (a,a) is in R.

*The relation is symmetric.

If a^2=b^2, then b^2=a^2.

*The relation is transitive.

If a^2=b^2 and b^2=c^2, then a^2=c^2.

*The relation is not antisymmretric.

(1,-1) and (-1,1) is in the relation but-1 doesn't equal 1.

c)

*The relation is reflexive.

a/a=1 for any a in the naturals.

*The relation is not symmetric.

Wile 4/2 is an integer, 2/4 is not.

*The relation is transitive.

If a/b=z and b/c=y where z and y are integers, then a=bz and b=cy.

This means a=cyz. This implies a/c=yz.

Since the product of integers is an integer, then (a,c) is in the relation provided (a,b) and (b,c) are in the relation.

*The relation is antisymmretric.

Assume (a,b) is an R. (Note: a,b are natural numbers.) This means a/b is an integer. This also means a is either greater than or equal to b. If b is less than a, then (b,a) is not in R. If a=b, then (b,a) is in R. (Note: b/a=1 since b=a)

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\bf Step-by-step~explanation:\\

\bf13.

We have to name a line that "contains" point P. This means that we have to find a line that passes through point P. We can see that line PS/n goes through point P.

\bf14.

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\bf 15.

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\bf 16.

We have to name a point that does not contain lines l, m, or n. Point W is a point that is not on any of the lines.

\bf 17.

Another name for line n is also line PS (add a line on the top of PS when writing this!) because point P and point S are two points on line n.

\bf 18.

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