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finlep [7]
4 years ago
5

Paul can install a 300 hundred square foot hardwood floor in 18 hours. Matt can install the same floor in 22 hours. How long wou

ld it take for Paul and Matt to install the floor working together?
Mathematics
1 answer:
Vadim26 [7]4 years ago
7 0
ANSWER

Paul and Matt will take
9.9 \: \: hours
to install the floor working together.

EXPLANATION

First, calculate Paul's rate, which is

= \frac{1}{18}

Next, calculate Matt's rate, which is,
= \frac{1}{22}

Let us assume that, Paul and Matt took
x \: hours
to do the work together.

Then, the rate of working together is,

= \frac{1}{x}

Now, add the individual rates and equate to the rate of working together to form an equation that will help us solve for x.

\frac{1}{18} + \frac{1}{22} = \frac{1}{x}

Simplify the left hand side

\frac{11 + 9}{198} = \frac{1}{x}

\frac{20}{198} = \frac{1}{x}

We reciprocate both sides to get (You could have also done cross multiplication).

\frac{198}{20} = \frac{x}{1}

x = 9.9 \: hours
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Answer:

3.6 liters

Step-by-step explanation:

So the ceiling an the walls, you are covering 5/6 surfaces.

Front  4.2*2.5= 10.5

Side    4*2.5=10

1 Top  4.2*4= 8.4

All together =28.9 M /8= 3.6 liters of paint would be needed

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In ΔMNO, o = 3.1 cm, ∠N=44° and ∠O=35°. Find the length of n, to the nearest 10th of a centimeter.
kirill115 [55]

Answer:

3.8

Step-by-step explanation:

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3 0
3 years ago
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Consider the circle where QR = 7 units and UT= 4 units. What is the value of X, if x represents the length of segment ST? A. 15
julia-pushkina [17]

Option C:

x = 6  units

Solution:

QR = 7 units, RS = 5 units, UT = 4 units and ST = x

<em>If two secants intersect outside a circle, the product of the secant segment and its external segment s equal to the product of the other secant segment and its external segment.</em>

⇒ SR × SQ =  ST × SU

⇒ 5 × (5 + 7) =  x × (x + 4)

⇒ 5 × 12 =  x² + 4x

⇒ 60 =  x² + 4x

Subtract 60 from both sides.

⇒ 0 = x² + 4x - 60

Switch the sides.

⇒ x² + 4x - 60 = 0

Factor this expression, we get

(x - 6)(x + 10) = 0

x - 6 = 0, x + 10 = 0

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Length cannot be in negative measures.

x = 6 units

Option C is the correct answer.

6 0
3 years ago
A student club holds a meeting. The predicate M(x) denotes whether person x came to the meeting on time. The predicate O(x) refe
Novay_Z [31]

Answer:

a) \exists \, x \in C : O(x) = 0

b) \{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}

c) \{ x \in C: M(x) = 1 \} = C

d) \{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C

e) \exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1

f) \exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0

Step-by-step explanation:

  • M(x) = 1 if the person x came to the meeting, and 0 otherwise.
  • O(x) = 1 if the person is an officer of the club and 0 otherwise.
  • D(x) = 1 if the person has paid hid/her club dues and 0 otherwise.

Lets also call C the set given by the members of the club. C is the domain of the functions M, O and D.

a) If someone is not an officer, the there should be at least one value x such that O(x) = 0. This can be expressed by logic expressions this way

\exists \, x \in C : O(x) = 0

b) If all the officers came on time to the meeting, then for a value x such that O(x) = 1, we also have that M(x) = 1. Thus, the set of officers of the Club is contained on the set of persons which came to the meeting on time, this can be written mathematically this way:

\{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}

c) If everyone was in time for the meeting, then C is equal to the set of persons who came to the meeting on time, or, equivalently, the values x such that M(x) = 1. We can write that this way:

\{ x \in C: M(x) = 1 \} = C

d) If everyone paid their dues or came on time to the meeting, then if we take the set of persons who came to the meeting on time and the set of the persons who paid their dues, then the union of the two sets should be the entire domain C, because otherwise there should be a person that didnt pay nor was it on time. This can be expressed logically this way:

\{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C

e) If at least one person paid their dues on time and came on time to the meeting, then there should be a value x on C such that M(x) and D(x) are both equal to 1. Therefore

\exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1

f) If there is an officer who did not come on time for the meeting, then there should be a value x in C such that O(x) = 1 (x is an officer), and M(x) = 0. As a result, we have

\exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0

I hope that works for you!

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4 years ago
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