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1 year ago

A system of equations can never have

Mathematics

2 answers:

Oduvanchick [21]1 year ago

6 0

Answer:

This shows that a system of equations may have one solution (a specific x,y-point), no solution at all,

explanation: solution (being all the solutions to the equation). You will never have a system with two or three solutions; it will always be one, none, or infinitely-many.

miskamm [114]1 year ago

4 0

More variables than equations

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A(c−b)=d solve for c

Rainbow [258]

Answer:

The solution to this equation could not be determined.

Step-by-step explanation:

Simplifying

a(c + -1b) = d

Reorder the terms:

a(-1b + c) = d

(-1b * a + c * a) = d

(-1ab + ac) = d

Solving

-1ab + ac = d

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Combine like terms: d + -1d = 0

-1ab + ac + -1d = 0

The solution to this equation could not be determined.

4 0

2 years ago

Please solve the question below 1

Fiesta28 [93]

Answer:

Step-by-step explanation:

Find the mean:

Day 1: 2

Day 2: 1

Day 3: 4

Day 4: 5

Day 5: 0

Day 6: 7

Mean = $\frac{2 + 1 + 4 + 5 + 0 + 7}{6} = 3.17$

This does not exceed a daily mean of 3.5 hours which was set by her mom as a guideline.

Therefore we can conclude that:

"Chawrdi complied with the guideline because the mean number of hours played is $\frac{2 + 1 + 4 + 5 + 0 + 7}{6} = 3.17$ .

8 0

2 years ago

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

exis [7]

The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

$\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1$

so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$