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Masja [62]
3 years ago
7

Consider this figure of a hexagon.

Mathematics
2 answers:
notsponge [240]3 years ago
6 0
A)
It looks like the [irregular] hexagon has 3 rectangles and 2 triangles within it.
So let's exclude the triangular corners on bottom left and top right for now.
First we have a large rectangle covering most of the upper left of the polygon. 20 ft × 7 ft = 140 sq.ft.
Now we have a rectangle on the bottom right. The width is 11 ft, so take the 7 away from that, 4 ft. × 14 ft. on bottom. 4 ft × 14 ft = 56 sq.ft.
The last small rectangle fits on the right between the 2 other rectangles. It is 24-20 on top/bottom × 7-6 right/left. 4 ft × 1 ft = 4 sq.ft.
Now for the triangles: bottom left is 11-7 × 24-14 = 4 ft × 10 ft. 1/2bh = 1/2×10×4 = 20 sq.ft.
Top right is 24-20 × 11-5 = 4 ft × 6 ft. 1/2bh = 1/2×4×6 = 12 sq.ft.

B)
Add them all together for the total area (A):
A = 140 + 56 + 4 + 20 + 12 = 140+60+32
= 232 sq.ft.

Hope that explains it well enough! ;)
just olya [345]3 years ago
3 0

Answer:

Step-by-step explanation:

(A)

1 big rectangle-2 corner triangles

(B)

24 * 11 = 264 square feet the triangles

11-7

= 4

24 -14

=10

1/2 * 4 * 10

= 20

24-20 by 11-6

=1/2 * 4 * 5 =10

=264 - (20+10)

= 264 - 30

=234 square feet

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Please help me with this
Tju [1.3M]

Answer:

Option 2: m∠1 = 147°, m∠2 = 80°, m∠3 = 148°

Step-by-step explanation:

Step 1: Consider triangle ABC from the picture attached below.

Lets find angle x

x + 47 + 33 = 180 (because all angles of a triangle are equal to 180°)

x = 100°

Angle x = Angle y = 100° (because vertically opposite angles are equal)

Step 2: Find angle 2

Angle 2 = 180 - angle x (because angle on a straight line is 180°)

Angle 2 = 180 - 100

Angle 2 = 80°

Step 3: Find angle z

48 + y + z = 180° (because all angles of a triangle are equal to 180°)

z = 32°

Angle 3 = 180 - angle z (because angle on a straight line is 180°)

Angle 3 = 180 - 32

Angle 3 = 148°

Step 4: Find angle 1

Angle 1 = 180 - 33 (because angle on a straight line is 180°)

Angle 1 = 147°

Therefore m∠1 = 147°, m∠2 = 80°, m∠3 = 148°

Option 2 is correct

!!

3 0
3 years ago
Help math..................​
mojhsa [17]

Answer:

6)C,120

7)B,60°

DOES THE ANSWER HELP YOU?

5 0
2 years ago
Question regarding logarithms.
Eddi Din [679]

5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}

\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}

\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}

6 0
3 years ago
Read 2 more answers
Help me with fractions. Please look at the picture before solving. Circle 1/5 of all the triangles. Mark X's on 2/3 of the trian
Irina-Kira [14]
1/5 are 3 triangles
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6 0
3 years ago
Read 2 more answers
Can someone please help me real quick?​
Ivahew [28]

Answer:

Big square - unshaded = shaded

((8+4)X(10+5))-(10X8)=100

Answer: 100

Step-by-step explanation:

8 0
2 years ago
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