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4 years ago

2x – 6y = 5

Mathematics

1 answer:

Juli2301 [7.4K]4 years ago

5 0

Answer:

The answer to your question is This system of equations can not be solved. They do not cross.

Step-by-step explanation:

2x - 6y = 5 ----- (I)

x - 3y = -12 ----- (II)

I.- Solve equation II for x

x = 3y - 12 ------ (III)

2.- Substitute equation III on equation I

2(3y - 12) - 6y = 5

3.- Expand

6y - 24 - 6y = 5

6y - 6y = 5 + 24

0 = 29

This system of equations can not be solved.

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algol13

<h3>Answer:</h3>

$\large\boxed{-1\frac{16}{25},\,\frac{6}{40},\,0.35,\,1\frac{3}{4}}$

<h3>Step-by-step explanation:</h3>

In this question, it's asking you to put the numbers that were given from <em>least </em>to <em>greatest.</em>

Our given numbers are:

$\frac{6}{40}$
0.35
$1\frac{3}{4}$
$-1\frac{16}{25}$

Now, lets sort them out.

We know that negative numbers would be the least. Sicne there's only one negative number, we would put that first because it's the least out of the numbers.

$\frac{6}{40}$ would go next. To make it easier, we can turn it into a decimal. $\frac{6}{40} = 0.15$ when you divide.

0.35 will go next. This would be bigger than 0.15, but lower than the next number.

$1\frac{3}{4}$ would go last, due to the fact that it's the greatest. $1\frac{3}{4}$ is the same as 1.75

When you put them in order, you should get $-1\frac{16}{25},\,\frac{6}{40},\,0.35,\,1\frac{3}{4}$

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>

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3 years ago

If a cylinder has a height of 12 cm and a radius of 4 cm, what do you do to find the volume of the cylinder?

jenyasd209 [6]

Answer:

V = 603.43 cm³

Step-by-step explanation:

V = πr²h

Where r = 4 cm, h = 12 cm

V = 22/7 × 4² × 12

V = 22/7 × 4 × 4 × 12

V = 4224/7

V = 603.43 cm³

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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

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

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To find the length of the side, will take the under root of the area. So, the side will be :

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4 years ago

The formula Y=X⁴-4X³+3X² is growing?

miskamm [114]

Answer:

The best way to know weather the formula y=x⁴-4x³+3x² is growing or not, is by graphing it.

As you can see in the attached picture:

For -inf<x< 0 the graph decreases.
For 0<x<0.634 the graph is growing
For 0,634<x<2.366 the graph decreases
For 2.366<x<+inf the graph is growing.

Therefore, the polynomial grows in the intervals stated before.

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3 years ago