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3 years ago

15

Jackson's kitten weighed 2 pounds 3 ounces. A month later the kitten weighed 56 ounces. How much did the kitten gain in the mont

h?

1 answer:

polet [3.4K]3 years ago

6 0

The kitten gained 21 ounces. 2 pounds is equal to 32 ounces, 32 + 3 = 35 and 56 - 35 = 21 ounces

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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

1 year ago

On a certain hot summers day , 293 people used the public swimming pool. Daily prices are $1.50 for children and $2.50 for adult

fredd [130]

C = children

A = adults

293 = c + a

1.50c + 2.50a = 676.50

We can use substitution to solve:

c + a = 293 subtract a to get c = 293 - a

Plug this into the second equation:

1.50(293 - a) + 2.50a = 676.50

439.5 - 1.50a + 2.50a = 676.50

439.5 + 1a = 676.50

a = 237

Substitute this into the first equation:

293 = c + 237

56 = c

8 0

3 years ago

The equation of a parabola is 1/32 (y−2)2=x−1 .

Andru [333]

The given equation of parabola is

$\frac{1}{32} (y-2)^2 = x-1$

Which can also be written as

$x = \frac{1}{32} (y-2)^2 +1$

Here vertex (h,k) is (1,2)

And value of a is

$a = \frac{1}{32}$

Formula of focus is

$(h+ \frac{1}{4a} , k)$

Substituting the values of h,k and a, we will get

$(1+ \frac{1}{4*(1/32) } , 2} = (1+ 8,2) = (9,2)$

Therefore the correct option is the last option .

6 0

3 years ago

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Which shows the image of quadrilateral ABCD after the transformation R0 90?

Arisa [49]

Answer:

second picture

Step-by-step explanation:

The rule to rotate a point counterclockwise about the origin is:

Or in words, we switch the coordinates of the point and change the sign of the y-coordinate.

We know from the first picture that the coordinates of our quadrilateral ABCD are A = (-1, 0), B = (0, -1), C = (-2, -3), and D = (-3 , -2), so let's apply the rotation rule to each one of those points:

Now we know that the coordinates of the quadrilateral after a rotation of 90° are A' = (0, 1), B' = (1, 0), C' = (3, -2), and D' = (2, -3), which corresponds to the second picture

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3 years ago

If x=-1, then which of the following equations makes a true statement? A. 4x+9= 20 B. -4x-5=-15 C. -3x+15=-22

AveGali [126]

Answer:

The only thing I can think of is that you miss-typed out the question. None of these work :/

8 0

3 years ago

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