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3 years ago

Calculate the surface area of a solid box in the shape of a cube with side length 10cm

Mathematics

1 answer:

iragen [17]3 years ago

7 0

Answer:

600 cm

Step-by-step explanation:

Surface Area of a box: SA=2lw+2lh+2hw

Since a box has equal side lengths, l = 10, w = 10, and h = 10.

SA = 2(10)(10) + 2(10)(10) + 2(10)(10)

SA = 600cm

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Can two vertical angles form a linear pair?

weqwewe [10]

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3 years ago

Help me from 6-9 (geometry)

elena55 [62]

Answer:

6) Two angles are called supplementary when their measures add up to 180 degrees. For example, two right angles are supplementary angles.

7) When the sum of two angles is equal to 90 degrees, they are called complementary angles. For example, 30 degrees and 60 degrees are complementary angles.

8) you have to plot the points on the graph and draw a line and label it with L at one point and M at another point.

9) x = 20

m < A = 65

m < B = 25

Step-by-step explanation:

3x + 5 + 2x - 15 = 90

3x + 2x + 5 - 15 = 90

5x - 10 = 90

5x - 10 + 10 = 90 + 10

5x = 100

5 / 5x = 100 / 5

x = 20

A = 3x + 5

A = 3(20) + 5

A = 60 + 5

A = 65

B = 2x - 15

B = 2(20) - 15

B = 40 - 15

B = 25

8 0

3 years ago

The vertices of a triangle ABC are A(7, 5), B(4, 2), and C(9, 2). What is measure of angle ABC? 30° 45° 56.31° 78.69°

GenaCL600 [577]

The measure of angle ABC is 45°

Explanation

Vertices of the triangle are: A(7, 5), B(4, 2), and C(9, 2)

According to the diagram below....

Length of the side BC (a) $=\sqrt{(4-9)^2+(2-2)^2}= \sqrt{25}= 5$

Length of the side AC (b) $= \sqrt{(7-9)^2 +(5-2)^2}= \sqrt{4+9}=\sqrt{13}$

Length of the side AB (c) $= \sqrt{(7-4)^2 +(5-2)^2} =\sqrt{9+9}=\sqrt{18}$

We need to find ∠ABC or ∠B . So using Cosine rule, we will get...

$cosB= \frac{a^2+c^2-b^2}{2ac} \\ \\ cos B= \frac{(5)^2+(\sqrt{18})^2-(\sqrt{13})^2}{2*5*\sqrt{18} }\\ \\ cosB= \frac{25+18-13}{10\sqrt{18}} =\frac{30}{10\sqrt{18}}=\frac{3}{\sqrt{18}}\\ \\ cosB=\frac{3}{3\sqrt{2}} =\frac{1}{\sqrt{2}}\\ \\ B= cos^-^1(\frac{1}{\sqrt{2}})= 45 degree$