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2 years ago

5

Drag each object to show whether cost is proportional to area in the situation represented. NO BITLY BECUSE YOU A SCAM PLS ANSWE

R FAST NO BITLY

1 answer:

Slav-nsk [51]2 years ago

6 0

What’s a bitly? Also try elephapnt

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3 cm and 9 mm converted completely into centimetres to one decimal place is ____cm. Someone please help me it would be greatly a

makvit [3.9K]

9mm in centimeters is 0.9

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2 years ago

(10 POINTS AND BRAINLIEST FOR BEST). Last month Ben spent $380 on food and $420 on personal items.

lions [1.4K]

4000 is his total monthly balance.

420 + 380 = 800

800/4000 = 0.20

0.20 = 20%

He spent 20% of his monthly income on food and personal items.

6 0

2 years ago

The system of equations below represents p, the number of photo files, and v, the number of video files Helena uploaded to an on

ss7ja [257]

P + v = 650....p = 650 - v
4p + 9v = 2925

4(650 - v) + 9v = 2925
2600 - 4v + 9v = 2925
-4v + 9v = 2925 - 2600
5v = 325
v = 325/5
v = 65.....video files

p + v = 650
p + 65 = 650
p = 650 - 65
p = 585....photo files

answer : Helena uploaded 9 times as many photo files as video files
65 * 9 = 585

5 0

2 years ago

Read 2 more answers

Unsure how to do this calculus, the book isn't explaining it well. Thanks

krok68 [10]

One way to capture the domain of integration is with the set

$D = \left\{(x,y) \mid 0 \le x \le 1 \text{ and } -x \le y \le 0\right\}$

Then we can write the double integral as the iterated integral

$\displaystyle \iint_D \cos(y+x) \, dA = \int_0^1 \int_{-x}^0 \cos(y+x) \, dy \, dx$

Compute the integral with respect to $y$ .

$\displaystyle \int_{-x}^0 \cos(y+x) \, dy = \sin(y+x)\bigg|_{y=-x}^{y=0} = \sin(0+x) - \sin(-x+x) = \sin(x)$

Compute the remaining integral.

$\displaystyle \int_0^1 \sin(x) \, dx = -\cos(x) \bigg|_{x=0}^{x=1} = -\cos(1) + \cos(0) = \boxed{1 - \cos(1)}$

We could also swap the order of integration variables by writing

$D = \left\{(x,y) \mid -1 \le y \le 0 \text{ and } -y \le x \le 1\right\}$

and

$\displaystyle \iint_D \cos(y+x) \, dA = \int_{-1}^0 \int_{-y}^1 \cos(y+x) \, dx\, dy$

and this would have led to the same result.

$\displaystyle \int_{-y}^1 \cos(y+x) \, dx = \sin(y+x)\bigg|_{x=-y}^{x=1} = \sin(y+1) - \sin(y-y) = \sin(y+1)$

$\displaystyle \int_{-1}^0 \sin(y+1) \, dy = -\cos(y+1)\bigg|_{y=-1}^{y=0} = -\cos(0+1) + \cos(-1+1) = 1 - \cos(1)$

7 0

8 months ago

Evaluate the expression for x = 3, y=1/3 , and z = 5. 12x−3y4x−z/

Nesterboy [21]

The answer is 19 :)

8 0

2 years ago