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3 years ago

Help me plzzzzzzzzzzzzzzzzzzzzz

Mathematics

2 answers:

IceJOKER [234]3 years ago

6 0

Answer:

$1.25

Step-by-step explanation:

10-2.50=7.5

7.5/6=1.25

hope this helps :3

Talja [164]3 years ago

4 0

Answer: $1.25

Step-by-step explanation:

10-total

(1 notebook)-2.50, 6 pens

10-2.50=7.50

7.50/6(pens)=1.25

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Use scientific notation to estimate the number of inches in 1,225 miles. Include all calculations in your final answer. 1 inch ≈

masya89 [10]

1225*(1.578*10^-5)=1.933*10^-2

1.578*10^-5= 0.00001578

1225*0.00001578=0.0193305

0.0193305=1.933*10^-2

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3 years ago

What is the standard deviation of the following set of data? 8, 12, 14, 8, 9, 5, 17, 19, 31, 8

Y_Kistochka [10]

5,8,9,12,14,17,19,31

in order

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3 years ago

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Two trains leave the same station at the same time, but are headed in opposite directions. One train travels at 70 mph and the o

AURORKA [14]

Answer:

4 hours

Step-by-step explanation:

I hope this helps :)

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4 years ago

HELP QUICK!!..................

Tema [17]

Answer:

164.6

Step-by-step explanation:

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4 0

3 years ago

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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