All categories

11 months ago

Find the area of the figure

Mathematics

1 answer:

stira [4]11 months ago

6 0

Answer:

36 in.

Step-by-step explanation:

find the area of the rectangle and triangle first

area of rectangle = length x height

aR = (6 + 6) x 5

aR = 12 x 5

aR = 60

area of triangle = base x height x 1/2

aT = (6 + 6) x 4 x 1/2

aT = 12 x 4 x 1/2

aT = 24

now, to find the area of the shaded area subtract the triangle from the rectangle

area = rectangle - triangle

a = 60 - 24

a = 36

You might be interested in

Question

Serhud [2]

Answer:

The value of k is 11.

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

2=(k-3)/(5-1)

2=(k-3)/4

2*4=k-3

8=k-3

k=8+3

k=11

6 0

2 years ago

What’s 2x squared times -1

oee [108]

Answer:

-1 ^2

follow me

3 0

2 years ago

The length and width of a certain rectangle are both decreased by 50%. If the length and width of the new rectangle are then inc

alina1380 [7]

51% less.

The length is decreased by 50%, which is the same as being multiplied by 0.5, then it's increased by 40% which is another multiplication by 1.4. So the overall change is 0.5 * 1.4 = 0.7, or the newest rectangle is 70% of the size of the original rectangle. And since the area increases by the square, the newest rectangle has 0.7 * 0.7 = 0.49 of the area as the original rectangle. So it has 51% less area than the original rectangle.

5 0

2 years ago

How do you round 3.99 to the nearest tenths?

Lemur [1.5K]

3.99 = > 3 + 0.99

first let's round up the 0.99, and the 9 rounds up to 1.0 so 0 + 1.0 = 1.

3 + 1 = 4, rounded up.

to the nearest tentth, 4.0.

the .0 simply means 0 tenths, since it's a whole 4.

8 0

2 years ago

This is solving differential equations by sepeartion of variables

Brums [2.3K]

$\mathrm dy+k(y-70)\,\mathrm dx=0$

Separating variables gives

$\dfrac{\mathrm dy}{y-70}=-k\,\mathrm dx$

Integrating both sides gives

$\displaystyle\int\frac{\mathrm dy}{y-70}=-k\int\mathrm dx$
$\ln|y-70|+C_1=-kx+C_2$
$\ln|y-70|=-kx+C$

Solving for $y$ gives

$e^{\ln|y-70|}=e^{-kx+C}$
$y-70=e^{-kx}e^C$
$y=Ce^{-kx}+70$

As $y(1)=140$ , you get

$140=Ce^{-k}+70$
$70=Ce^{-k}$
$C=70e^k$

so the particular solution to this ODE is

$y=70e^ke^{-kx}+70$
$y=70e^{k(1-x)}+70$

Unless you have any more information, you're done.

5 0

2 years ago