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3 years ago

Create an equivalent expression for 9(5x - 4) by using the distributive property.

Mathematics

1 answer:

Bogdan [553]3 years ago

4 0

Answer:

3(15x-12)

Step-by-step explanation:

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Add the following a2b - ab2 + 8ab and 2a2b - ab2

Licemer1 [7]

Step-by-step explanation:

this is your answer............

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3 years ago

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6x-3=3x+12 what is the solution to the equation?

Alona [7]

Answer:

x=5

Step-by-step explanation:

add 3 to both sides of the equation: -3+3=0 12+3=15 simplified: 6x=3x+15 then subtract 3x from both sides of the equation. simplified: 3x=15

then divide both sides by 3. simplified: x=5

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3 years ago

Hello please help me

lozanna [386]

Answer:

Step-by-step explanation:

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3 years ago

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Please help me (30 points) 1. A recipe calls for 2 1/3 cups of flour, 3/4 cups of white sugar, and 1/3 cups of brown sugar. The

Katena32 [7]

Lets find how much flour we need for just one serving.

2 2/3 ÷ 6

8/3 ÷ 6

8/3 * 1/6

8/18

4/9 cup of flour per serving

If we triple the recipe (6 * 3 = 18), we will have a total of 18 servings. Multiply the amount of flour we got for one serving to 18 servings.

4/9 * 18/1

72/9

8 cups of flour for 18 servings

Best of Luck!

4 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