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3 years ago

Computer table measures 10ft by 17ft. Cost of table is $16 per sq ft. How much will the table cost.

1 answer:

alexandr1967 [171]3 years ago

8 0

Hello! First things first, we need to find the area of the table. Just by doing some basic math, 10 * 17 is 170. The area of the computer table is 170 square feet. The table costs $16 per square feet. To find the total cost, all you have to do is multiply the area of the table by the cost per square foot. 170 * 16 is 2,720. The computer table will cost $2,720.

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In the diagram below of a parallelogram STUV, SV=x+3, VU 2x-1, and TU=4x-3. What is the length of SV?

Katena32 [7]

Answer:

$SV=5\ units$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

In a parallelogram opposites sides are parallel and congruent

In this problem

$SV=TU$ ---> by opposite sides

substitute the given values

$x+3=4x-3$

solve for x

$4x-x=3+3\\3x=6\\x=2$

Find the length of SV

$SV=x+3$

substitute the value of x

$SV=2+3=5\ units$

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3 years ago

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2 years ago

Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:

professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is $\{1,2,3,4\}$ .

a) We need to find a relation R reflexive and transitive that contain the relation $R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}$

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

1R4 and 4R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
4R1 and 1R2, then 4 must be related with 2.

Therefore $\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\}$ is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

since 1R2, then 2 must be related with 1.
since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

1R2 and 2R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
2R1 and 1R4, then 2 must be related with 4.
4R1 and 1R2, then 4 must be related with 2.
2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

1 must be related with 1,

2 must be related with 2,

3 must be related with 3,

4 must be related with 4

For be symmetric

since 1R2, 2 must be related with 1,
since 1R4, 4 must be related with 1.

For be transitive

Since 4R1 and 1R2, 4 must be related with 2,
since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

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Answer:

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Step-by-step explanation:

Among digits, 10% of 10 digits is a single digit. Any answer choice that suggests more than one digit be used for "yes" is not appropriate. A good choice is ...

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Answer:

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Step-by-step explanation:

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