All categories

3 years ago

I need help!!!!!!!!!!

Mathematics

1 answer:

Andru [333]3 years ago

4 0

Answer:

Yes

Step-by-step explanation:

The points will be on the same line if they have the same slope. Calculate the slope of each pair. If the slopes are the same, then they lie on the same line.

$m = \frac{y_2-y_1}{x_2-x_1} = \frac{3-2}{1-4} = \frac{1}{-3} = -\frac{1}{3}$

$m = \frac{y_2-y_1}{x_2-x_1} = \frac{4-2}{-2-4} = \frac{2}{-6} = -\frac{1}{3}$

$m = \frac{y_2-y_1}{x_2-x_1} = \frac{4-3}{-2-1} = \frac{1}{-3} = -\frac{1}{3}$

The slope is the same.

You might be interested in

What is 10 divided by 5/12

Svetllana [295]

The answer is 24 now goodluck with your homework

3 0

3 years ago

Read 2 more answers

Please answer correctly !!!! Will mark brainliest !!!!!!

Musya8 [376]

Answer:

h(r) = 6(r+1)+10

Step-by-step explanation:

Just solve for q. Because that's the output. h(r) = q

3 0

3 years ago

Read 2 more answers

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

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

5 0

3 years ago

8. A carpenter cut the top section of a window frame with a 37 degree angle on each end. The side pieces each have 50 degree ang

marshall27 [118]

Answer:

I will assume that the end of the top section if it were cut at 0 degrees would be cut at right angles to the length of the section.

Joining a top section with a 37 degree end angle to a side piece with a 50 degree end angle will result in the length of the side piece being at an angle of

37 + 50 = 87 degrees to each other.

The same will be true for the other side.

If the sides were parallel, then going up one side, across the top and then down the other side will result in a change of direction of 180 degrees. However, in this frame the direction turns 87 degrees at the first top corner and another 87 degrees at the second top corner. This is a total of

87 + 87 = 174 degrees instead of 180 degrees.

The two sides diverge at an angle of

180 - 174 = 6 degrees from each other as they extend downwards from the top of the frame

Step-by-step explanation:

4 0

3 years ago

A field with an area of 2880 ft^2 has the shape of a rectangle, whose length is 12 ft larger than its width. Find the dimensions

tiny-mole [99]

ANSWER

w=48,l=60

EXPLANATION

Let the width of the field be, w , then,