All categories

3 years ago

Answer right then you get brainliest

Mathematics

1 answer:

tiny-mole [99]3 years ago

7 0

It's simple. The base of the rectangle is equal to the base of the big triangle. The height of the rettangle is equal to the base of the small triangle.

First you can calculate the area of the rectangle.

Ar = b*h = 5*8 = 40 in^2

Then you calculate the area of the two triangles

Small triangle area = (b*h)/2 = (6*5)/2 = 30/2 = 15 in^2

Big triangle area = (b*h)/2 = (8*7)/2 = 56/2 = 28 in^2

Sum the area of all polygons.

40+15+28 = 83 in^2

Answer C

You might be interested in

8, 9, and 10, please?? tell me how you got it, I'm so confused

Fittoniya [83]

Pretty sure mean for all of them because that is the calculated average.

5 0

3 years ago

Calculate all four second-order partial derivatives and confirm that the mixed partials are equal. f(x,y)= 2e^xy

nadya68 [22]

ANSWER TO QUESTION 1

The given function is

$f(x,y)=2 {e}^{xy}$

The partial derivative of f with respect to x means we are treating y as a constant. The first derivative is

$f_{x} = 2y {e}^{xy}$

and the second derivative with respect to x is,

$f_{xx} = 2 {y}^{2} {e}^{xy}$

ANSWER TO QUESTION 2

The given function is

$f(x,y)=2 {e}^{xy}$

The partial derivative of f with respect to y means we are treating x as a constant. The first derivative is

$f_{y} = 2x{e}^{xy}$

and the second derivative with respect to y is

$f_{yy} = 2 {x}^{2} {e}^{xy}$

ANSWER TO QUESTION 3

Our first mixed partial is

$f_{xy}$

We need to differentiate
$f_{x} = 2y {e}^{xy}$
again. But this time with respect to y.

Since this is a product of two functions of y, we apply the product rule of differentiation to obtain,

$f_{xy} = 2y( {e}^{xy})' + ({e}^{xy})(2y)'$

$f_{xy} = 2xy {e}^{xy} + 2{e}^{xy}$

ANSWER TO QUESTION 4

The second mixed partial is

$f_{yx}$

We need to differentiate
$f_{y} = 2x{e}^{xy}$

again. But this time with respect to x.

Since this is a product of two functions of x, we apply the product rule of differentiation to obtain,

$f_{yx} = 2x({e}^{xy})' + ({e}^{xy})(2x)'$

$f_{yx} = 2xy {e}^{xy} + 2{e}^{xy}$

Hence,

$f_{xy} = 2xy {e}^{xy} + 2{e}^{xy} =f_{yx}$

4 0

3 years ago

Please respond and help

jeyben [28]

Answer:

Step-by-step explanation:

7 0

3 years ago

Which is not an equation of the line going through (6, 7) and (2,-1)?

atroni [7]

Answer:

B. y-1=2(x+2)

Step-by-step explanation:

be the points (6,7) and (2,-1)

replacing 7-1=2(6+2) → 6≠16 and -1-1=2(2+2) → -2≠8

for all equations the equalities are maintained

3 0

3 years ago

The median weight of 21 dogs in a pet store is 14 pounds. Which action could CHANGE the median? A) the store gets a new 14-pound

mylen [45]

Answer:

B. the store gets a new 40-pound dog

Step-by-step explanation:

5 0

3 years ago