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3 years ago

Determine whether each equation represents a linear or non linear function, help me plzz

Mathematics

1 answer:

Ksju [112]3 years ago

5 0

Answer:

y=3x - Linear Function

y=4x-1 - Linear Function

y=x²+1 - Non-Linear Function

y=0.75x+2 - Linear Function

y=0.5x³ - Non-Linear Function

(For a function to be linear, both x & y must equal 1.)

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3 years ago

The zeros of this using quadratic equation

aleksandr82 [10.1K]

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4 years ago

What is the median of the ages shown in this stem-and-leaf plot? 59.5 58.5 59.0 58.0 Stem and leaf plot titled Ages of college p

AlexFokin [52]

Answer:

Median = 59.0

Step-by-step explanation:

According to the given data:

The data represents the signatures collected by different volunteers are as following in order from least to the greatest:

30, 32, 41, 44, 44, 45, 53, 55, 58, <u>59</u>, 62, 62, 62, 65, 66, 68, 75, 75, 76

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The median of a data set is the middle data point.

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<u>So, the median is 59</u>

8 0

3 years ago

Read 2 more answers

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

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

3 0

2 years ago