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harina [27]
2 years ago
12

Help me with this please.

Mathematics
1 answer:
andriy [413]2 years ago
4 0
1. bc/3 + a/xy + 3/ab
= b^2cxya + 3a^2b + 9xy/3xyab

2. (5/3y) + 2y - 12
=11/3y - 2

3. ?

Hope this helps and right! :)) Good luck
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2p2 -15p + 25 =<br><br> (p - 5)(2p - 5)<br> (p - 5)(2p + 5)<br> (p + 5)(2p - 5)<br> (2p - 15)(p + 5)
Anarel [89]
I think it's the first one
7 0
3 years ago
Read 2 more answers
State the domain and range of the relation. Determine whether the relation represents a function.
dangina [55]

Answer:

Domain {-2,0,2}

Range {-2,0,2}

Relation is a Function

Step-by-step explanation:

We are given a relation:

{ (-2,-2) , (0,0) , (2,2) }

Domain can be defined as the all possible values of x for a relation. It is considered as a set of all first values of the ordered pairs of a given relation.

Domain of the given relation is {-2,0,2}

Range can be defined as all possible value of y which corresponds to the values of x in the domain. It is considered as a set of all second values of the ordered pairs of a given relation.

Range of the given relation is {-2,0,2}

A relation is a function if only there is one value of y for each value of x. If in the set of ordered pair of the relation, the value of x gets repeated, then the relation is not a function.

As no values of x are getting repeated, the relation is a function.

4 0
3 years ago
4 -2x plus 5 when x =4
Bess [88]

Answer:

1

Step-by-step explanation:

f(4)=4-2x+5

4-2(4)+5

4-8+5

1

6 0
2 years ago
Read 2 more answers
8 less than the product of 5 and -4
Karolina [17]
The answer would be:
5(-4) - 8
-20 - 8
-28
4 0
3 years ago
The base of an auditorium is in the form of an eclipse 200 feet long and 100 feet wide a pin drop near one focus can clearly be
dedylja [7]

Answer:

Let the coordinate of focus be (\pm c , 0)

As per the statement: The base of an auditorium is in the form of an eclipse 200 feet long and 100 feet wide.

⇒Length of Major axis=base of an auditorium = 200 feet and Length of a minor axis=wide of a auditorium = 100 ft

Semi-major axis (a) = 100 ft and

semi-minor axis(b) = 50 ft

Then, by an equation:

c^2 = a^2-b^2

Solve for c:

Substitute the given values we have;

c^2=(100)^2-(50)^2

Simplify:

c^2 = 7500

or

c=\sqrt{7500} = 86.6025404 ft

Distance between the foci is,  2c = 2 \cdot 86.6025404 = 173.205081

Therefore, the distance between the foci to the nearest 10th of a foot is, 173.2 ft

6 0
3 years ago
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