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2 years ago

Solve for y: 4x + 5y = 12

Mathematics

1 answer:

Kryger [21]2 years ago

3 0

Answer:

y = (12 - 4x)/5

Step-by-step explanation:

5y = 12 - 4x

y = 12/5 - 4x/5

y = (12 - 4x)/5

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2 years ago

Read 2 more answers

Helppppppp show the steps to your answer

Studentka2010 [4]

The numeric value of the expression -a² - 2bc - |c| for a = -3, b = -5 and c = 2 is of 9.

<h3>How to find the numeric value of an expression?</h3>

The numeric value of an expression is found replacing each letter by it's attributed value.

In this problem, the expression is:

-a² - 2bc - |c|

The attributed values are:

a = -3, b = -5 and c = 2

Hence the numeric value will be given by:

-a² - 2bc - |c| = -(-3)² - 2(-5)(2) - |2| = -(9) + 20 - 2 = -9 + 18 = 9.

More can be learned about the numeric value of an expression at brainly.com/question/14556096

#SPJ1

4 0

1 year 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

1 year ago