All categories

2 years ago

Find the area of the shaded region.pls do this with explanation.

Mathematics

2 answers:

vladimir1956 [14]2 years ago

6 0

Answer:

first find the area of the big rectangle = 12×14=168

area of the small unshaded rectangle =2×5×6=60

area of the shaded region = 168-60=108

coldgirl [10]2 years ago

5 0

ANSWER

My answer is in the photos above

You might be interested in

What is negative 1 minus negative 1

lana [24]

-1 - -1.
-1 + 1.
0.

The answer is 0.

3 0

3 years ago

Read 2 more answers

For all the following, which ones represent y as a function of x, and how do you know?

earnstyle [38]

Only 25 and 28 are functions.

A function must have only one y value per x value. For 26, the y value of 0 and multiple others have two different y values. For 27, all x values less than 0 have multiple y values.

4 0

3 years ago

Find the solutions for a triangle with a = 16, c =12, and B = 63º.

maks197457 [2]

Answer:

b. A = 71.6°; C = 45.40°; b =15.0

Step-by-step explanation:

The missing values can be found with the help of the Law of Cosine and properties of triangles:

Side b (Law of Cosine)

$b = \sqrt{a^{2}+c^{2}-2\cdot a \cdot c \cdot \cos B}$

$b = \sqrt{16^{2}+12^{2}-2\cdot (16)\cdot (12) \cdot \cos 63^{\circ}}$

$b \approx 15.022$

Angle A (Law of Cosine)

$\cos A = -\frac{a^{2} - b^{2}-c^{2}}{2\cdot b \cdot c}$

$\cos A = - \frac{16^{2}-15.022^{2}-12^{2}}{2\cdot (15.022)\cdot (12)}$

$\cos A = 0.315$

$A= \cos^{-1} 0.315$

$A \approx 71.639^{\circ}$

Angle C (Sum of internal angles in triangles)

$C = 180^{\circ} - 63^{\circ} - 71.639^{\circ}$

$C = 45.361^{\circ}$

Hence, the right answer is B.

6 0

3 years ago

In the graph below, Point A represents Owen's house, Point B represents David's house and Point C represents the school. Who liv

ycow [4]

Answer:

• David

• 4 miles

Explanation:

In the graph:

The given locations are:

• Owen's House, A(11,3)

• David's House, B(15,13)

• School, C(3,18)

We determine both Owen's and David's distance from the school using the distance formula.

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Owen's distance from school (AC)

$\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}$

David's distance from school (BC)

$\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}$

We see from the calculations that David lives closer to the school, and by 4 miles.

The graph below is attached for further understanding: