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In-s [12.5K]
3 years ago
11

You are mailing a package that weighs 5 pounds, and sending it first class. The post office charges $0.44 for the first ounce, a

nd charges $0.20 for each additional ounce. How much is the total cost to mail this package?
Mathematics
1 answer:
Masja [62]3 years ago
4 0

Step-by-step explanation:

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Marlo had $35 but then spent $5​m
Step2247 [10]

Answer:

$30

Step-by-step explanation:

All you do is subtract the 5 dollars that he spent from the 35 dollars he already had and you get $30!

6 0
3 years ago
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A cube has side lengths of 2 m. What is the surface area of the cube? Enter your answer in the box.
k0ka [10]
24m^2 is the answer

Hoped that I helped you.
4 0
3 years ago
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If 50 grams of sweats cost $2.10,find the cost of 380 of sweats,giving your answer correct to the nearest 5 cent
Nezavi [6.7K]

Answer:

$15.95

Step-by-step explanation:

cost of 1 gram of sweet = $2.10 / 50 = 0.042

Cost of 380 g = 380 x 0.042 = $15.96

to the nearest 5 cent = 15.95

6 0
3 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}

3 0
2 years ago
Which of the following equations could be used to solve the given equation?
Pepsi [2]
If we simplify like terms on left and right sides we gwt

16x + 9 =  4x

Its B
8 0
3 years ago
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