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1 year ago

Select all the statements that best describe the table below.

Mathematics

1 answer:

guapka [62]1 year ago

7 0

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The Correct statements are :

The rate of change is not constant

The relationship is non linear

Independent variable is Age

Dependent variable is selling price

It is a negative relationship

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Andy takes an average of 12 breaths per minute About how many breaths did Andy take by the time he turned 13 years old?

Diano4ka-milaya [45]

6832800 breathes in 13 years exactly

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2 years ago

Saffron is a powder made from crocus flowers and is used in the manufacturing of perfume. Some 8,000 crocus flowers are required

Aneli [31]

16/2=8
we need 8 times more flowers
8x8,000=64,000
64 thousand flowers
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2 years ago

Read 2 more answers

I need help! who ever get this right and explain nicely then I make you brainlist.

Lelechka [254]

Answer:

1. 64

2. 72 square feet

3. 31.5 square feet

Step-by-step explanation:

For the first one use the equation A=bh.

Plug in your values: A=12.8*5.

Solve: A=64

For the second one multiply 8*3 which gives you 24. Next multiply that by 3 to give you 72 square feet.

For the third one multiply 9 by 3.5 to get 31.5 square feet.

5 0

2 years ago

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

A sign on Corey's house says it was built in MDCCCXXX. What year is this?

Bad White [126]

It is approximately 1830.

M=1000

D=500

C=100

X=10

7 0

2 years ago