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Effectus [21]
3 years ago
9

Is every linear relationship a direct variation. Explain please

Mathematics
1 answer:
pychu [463]3 years ago
3 0
Direct variation: y=kx where k is some number thats not 0

that means if x is 2, y is 2k. if x is 1, y is k. if x is 0, y is 0.

notice: no matter what k is, when you put 0 in for x, y will always be 0. that means that a direct variation always goes through the origin, a.k.a. (0,0).

do all lines go through the origin? nah. a line is y=mx+b, where m and b are some numbers. a line is only a direct variation if b is 0. this makes sense because if b is 0, then the y-intercept is 0, so the line goes through the origin.

in conclusion, no, not all linear relationships are direct varations, but all direct variations are linear relationships.
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Consider the differential equation x^2 y''-xy'-3y=0. If y1=x3 is one solution use redution of order formula to find a second lin
Anastasy [175]

Suppose y_2(x)=y_1(x)v(x) is another solution. Then

\begin{cases}y_2=vx^3\\{y_2}'=v'x^3+3vx^2//{y_2}''=v''x^3+6v'x^2+6vx\end{cases}

Substituting these derivatives into the ODE gives

x^2(v''x^3+6v'x^2+6vx)-x(v'x^3+3vx^2)-3vx^3=0

x^5v''+5x^4v'=0

Let u(x)=v'(x), so that

\begin{cases}u=v'\\u'=v''\end{cases}

Then the ODE becomes

x^5u'+5x^4u=0

and we can condense the left hand side as a derivative of a product,

\dfrac{\mathrm d}{\mathrm dx}[x^5u]=0

Integrate both sides with respect to x:

\displaystyle\int\frac{\mathrm d}{\mathrm dx}[x^5u]\,\mathrm dx=C

x^5u=C\implies u=Cx^{-5}

Solve for v:

v'=Cx^{-5}\implies v=-\dfrac{C_1}4x^{-4}+C_2

Solve for y_2:

\dfrac{y_2}{x^3}=-\dfrac{C_1}4x^{-4}+C_2\implies y_2=C_2x^3-\dfrac{C_1}{4x}

So another linearly independent solution is y_2=\dfrac1x.

3 0
3 years ago
Ava, Kate, and Jin bring all the food to a picnic. Ava brings 34, Kate brings 16, and Jin brings 112. Enter a numerical expressi
juin [17]

Answer:

Step-by-step explanation:

Given that:

Ava bring 34 food to the picnic

Kate bring 16 food to the picnic

JIn brings 112 food to the picnic

To enter these digits by applying equivalent fractions with a denominator of 12;

Their respective numerical expression is:

Ava = \dfrac{34}{12}

Ava brought 2 \dfrac{5}{6} food to the picnic

Kate = \dfrac{16}{12}

Kate brought 1 \dfrac{1}{3} food to the picnic

Jin = \dfrac{112}{12}

Jin brought 9 \dfrac{1}{3} food to the picnic

6 0
3 years ago
Read 2 more answers
assume that mrs.giang's car travels 14 miles on each gallon of gas. if she travels to visit her niece who lives 133 miles away,
zalisa [80]

Answer:

9.5 gallons of gas

Step-by-step explanation:

Given that:

Each gallon gives 14 miles

Mrs giang wants to travel 133 miles

So

gallons she would need  = 133 / 14 miles

gallons she would need = 9.5 gallons

So Mrs.giang will need 9.5 gallons of gas to visit her niece.

i hope it will help you!

8 0
4 years ago
Which describes the correlation shown in the scatterplot?
Alex_Xolod [135]
Answer: C) There is no positive or negative correlation

A positive correlation is one where the values are increasing, and the line of dots are clearly going down to up.

A negative correlation is the opposite. The values are decreasing, and the line goes from up to down.

This correlation is one that is neither positive or negative. It is a straight line where the values neither increase nor decrease.

Hope this helps comment below for more questions :)
8 0
3 years ago
Read 2 more answers
Fill in the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions: 2a
Vitek1552 [10]

Answer:

2ab - 6a + 5b - 15

Step-by-step explanation:

Given

2ab - 6a + 5b + \_

Required

Fill in the gap to produce the product of linear expressions

2ab - 6a + 5b + \_

Split to 2

(2ab - 6a) + (5b + \_)

Factorize the first bracket

2a(b - 3) + (5b + \_)

Represent the _ with X

2a(b - 3) + (5b + X)

Factorize the second bracket

2a(b - 3) + 5(b + \frac{X}{5})

To result in a linear expression, then the following condition must be satisfied;

b - 3 = b + \frac{X}{5}

Subtract b from both sides

b - b- 3 = b - b+ \frac{X}{5}

- 3 = \frac{X}{5}

Multiply both sides by 5

- 3 * 5 = \frac{X}{5} * 5

X = -15

Substitute -15 for X in 2a(b - 3) + 5(b + \frac{X}{5})

2a(b - 3) + 5(b + \frac{-15}{5})

2a(b - 3) + 5(b - \frac{15}{5})

2a(b - 3) + 5(b - 3)

(2a + 5)(b - 3)

The two linear expressions are (2a+ 5) and (b - 3)

Their product will result in 2ab - 6a + 5b - 15

<em>Hence, the constant is -15</em>

3 0
3 years ago
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