All categories

3 years ago

Two times the sum of a number and three is not equal to eight

Mathematics

1 answer:

jarptica [38.1K]3 years ago

8 0

Could be the sum is 1 and then 2 times 2 which equals 4 then 5 then plus 3 it is 8 so wrong

You might be interested in

How do I solve for d if d/M-3=R

nignag [31]

Answer:

d = (R+3)(M)

Step-by-step explanation:

d/M - 3 = R

Add 3 both sides

d/M = R+3

Multiply M both sides

d = (R+3)(M)

Hope this helps!

5 0

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

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

5 0

2 years ago

Add -2x^4+x^2-x-9 and x^4-x^3-5x+3

eimsori [14]

Answer:

Your solution would be -1x^4 - 1x^3 + x^2 - 6x - 6.

Step-by-step explanation:

5 0

2 years ago

Write an equation that passes through a pair of points

Phoenix [80]

Answer:

C. $y = -x + 2$

Step-by-step explanation:

First, find the slope, m, of the line that passes through the two given points.

Slope = $\frac{rise}{run} = \frac{-2}{2} = -\frac{1}{1} = -1$

Next, determine the y-intercept, b.

The y-intercept is where the line intercepts the y-axis. The line intercepts the y-axis at y = 2. Therefore, b = 2.

Now, to get an equation of the line, substitute m = -1, and b = 2 in $y = mx + b$ .

✅The equation would be:

$y = -1x + 2$

$y = -x + 2$

3 0

2 years ago

HELP ASAP PLEASE ASAP PLEASE

Alexandra [31]

Answer:

it should be B

mark me as the brainliest please

3 0

3 years ago