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artcher [175]
2 years ago
13

The cost per week of running a boarding house is partly constant and partly

Mathematics
1 answer:
patriot [66]2 years ago
6 0

Answer:

the cost of running the boarding

house for 600 students is N61,000

Step-by-step explanation:

Let C represents cost

K1 represents first constant

K2 represents second constant

C= k1+k2n

3500 = k1 + 25 k2............. Eqn(1)

6000= k1 + 50 k2 .............. Eqn(2)

Subtract eqn(1) from eqn(2)

2500= 25k2

K2= 2500/25

K2= 100

To get k1 from eqn(1)

3500 = k1 + 25 k2

Substitute the value of k2

3500 = k1 + 25 (100)

3500= k1 +2500

K1= 3500- 2500

K1= 1000

The equation connecting them;

C= 1000+ 100n

The cost of running the boarding

house for 600 students is

n= 600

C= 1000+ 100(600)

C= 1000+60000

C= N 61,0000

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C=6.28cm
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4 0
3 years ago
If θ is an angle in standard position that passes through (-3, 4), find sin2θ.
julia-pushkina [17]

Answer:

Correct answer: sin 2Θ = - 24/25

Step-by-step explanation:

If under the standard position you think that the first arm (side) belongs to the positive direction of the x axis and the second one passes through a given point then it is:

if we form a right triangle with sides 3 and 4 then the hypotenuse is 5.

sin Θ = 4/5 and cos Θ = - 3/5

we know that the formula for double the value of the angle is:

sin 2Θ = 2 sinΘ cosΘ = 2 · 4/5 · ( - 3/5) = - 24/25

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God is with you!!!

8 0
3 years ago
Determine whether each of the binary relation R defined on the given sets A is reflexive, symmetric, antisymmetric, or transitiv
ki77a [65]

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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2 years ago
How do i evaluate 16^0.25 ?
KengaRu [80]
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3 years ago
I am a factor of 18.the other factor is 9.what number am i?
Liula [17]
The answer to this question is 2

3 0
3 years ago
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