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3 years ago

10

The area of the trapezoid is 660 sq cm. Given the sides of 26 cm each and bases of 42 cm and 24 cm, find the height of the trape

zoid.

1 answer:

adell [148]3 years ago

3 0

Hand sham. Your answer is the number below
7392.3

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Change y-7=m(x-4) to standard form

Ivan

Answer:

mx -y = 4m -7

Step-by-step explanation:

Standard form is ...

ax +by = c

where a, b, c are mutually prime integers and a > 0.

If we assume m > 0, then we need to collect the variable terms on the right side of the equation, so the coefficient of x will be positive.

y -7 = mx -4m . . . . eliminate parentheses

-7 = mx -y -4m . . . . subtract y

4m -7 = mx -y . . . . . add 4m

mx -y = 4m -7 . . . . . . standard form

4 0

2 years ago

You can buy DVDs at a local store for $15.49 each. You can buy them at an online store for $13.99 each plus $6 for shippimg. How

NeTakaya

15.49x = 13.99x + 6
15.49x - 13.99x = 6
1.50x = 6
x = 6 / 1.50
x = 4 <=== u can buy 4 DVD's for the same amount

15.49 * 4 = 61.96
13.99(4) + 6 = 61.96

6 0

3 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

1 year ago

Divide using synthetic division (3y^4-13y^3+12y^2-33y+9)/(y-4)

tatiyna

3y³-y²+8y-1+5/y-4. is that unless you want the answer smaller

3 0

3 years ago

Help? Please? Thank you!

sveta [45]

Answer:

independent: miles

dependent: minutes/hours

Step-by-step explanation:

can I see the other options for the rest?

3 0

2 years ago