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3 years ago

2+2=4-1?

Mathematics

2 answers:

astra-53 [7]3 years ago

8 0

BTS is sooooo trash .....

evablogger [386]3 years ago

7 0

Answer:

omggg BUTTER IS SO GOOODDD

I M LOVING ITT!!!!

#BTSBUTTER

#BTS

#ARMY

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The equations of four lines are given. identify which ones are perpendicular.

andriy [413]

Answer:

line 2: y=1/5x-3

Step-by-step explanation:

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2 years ago

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Write an equation which represents c, the total cost of n pizzas purchased.

bezimeni [28]

C= total $ cost
n= number of pizzas purchased
x= $ cost per pizza

C= x * n
C= xn

If you know the cost per pizza and the number of pizzas purchased, plug those numbers in the equation above to find the total cost.

EXAMPLE: If each pizza cost $10 and 5 were purchased, the equation would be: C= ($10)(5)= $50 total

ANSWER: C= xn

Hope this helps! :)

4 0

4 years ago

Please please help with math!

Sergio [31]

The answer is B. Hope it helped

6 0

3 years ago

A bag has four green marbles and eight red marbles. Half of the red marbles are made of glass. A marble is selected at random fr

Ratling [72]

Answer:

4/12 1/3

Step-by-step explanation:

4/4/4 so u have a 4/12 or 1/3 chance

6 0

3 years ago

Calculus help please, will give brainiest

Andrew [12]

The most convenient way to capture $D$ is with the parameterization

$D = \left\{(x,y) \mid -1 \le y \le 3 \text{ and } -\dfrac{y+1}2 \le x \le y+1\right\}$

so we only need one iterated integral.

$\displaystyle \iint_D e^{x-y} \, dA = \int_{-1}^3 \int_{-(y+1)/2}^{y+1} e^{x-y} \, dx \, dy$

Compute the integral with respect to $x$ .

$\displaystyle \int_{-(y+1)/2}^{y+1} e^{x-y} \, dx = e^{x-y} \bigg|_{x=-(y+1)/2}^{x=y+1} = e^{(y+1)-y} - e^{-(y+1)/2-y} = e - e^{-(3y+1)/2}$

Compute the remaining integral.

$\displaystyle \int_{-1}^3 \left(e - e^{-(3y+1)/2}\right) \, dy = \left(ey + \frac23 e^{-(3y+1)/2}\right)\bigg|_{y=-1}^{y=3} \\\\ ~~~~~~~~ = \left(3e + \frac23 e^{-(9+1)/2}\right) - \left(-e + \frac23 e^{-(-3+1)/2}\right) \\\\ ~~~~~~~~ = \boxed{\frac{10e}3 + \frac2{3e^5}}$

If we had chosen the opposite order of variables, we would have used

$D = \left\{(x,y) \mid -2 \le x \le 4 \text{ and } \max\left(-2x-1,x-1\right)\le y\le3\right\}$

where

$\max(-2x-1, x-1) = \begin{cases} -2x-1 & \text{when } x$

so we would have needed two iterated integrals,

$\displaystyle \iint_D e^{x-y} \, dA = \int_{-2}^0 \int_{-2x-1}^3 e^{x-y} \, dy \, dx + \int_0^4 \int_{x-1}^3 e^{x-y} \, dy \, dx$

3 0

2 years ago