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dem82 [27]
3 years ago
8

Find area AND perimeter of the shape

Mathematics
2 answers:
den301095 [7]3 years ago
8 0

Step-by-step explanation:

for the perimeter...

18-6=12

that's to get the slant side

perimeter=10+6+12+12+20

=60

for the area I'm not sure how to solve it sorry!!

Rashid [163]3 years ago
6 0

Answer:

10 x 6 = 60

20 + 10 = 30, 30 x 12 x 1/2 = 180

180 + 60 = 240 cm2 is your area

6 + 6 = 12, 12 + 10 = 22, 22 + 20 = 42

Now lets find hypotenuse:

10 - 20 = 10, 10 divided by 2 = 5

5^2 + 12^2

25 + 144 = 169

Square root of 169 is 13

Add perimeters:

42 + 13 = 55 cm is your perimeter.

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Simplify the expression 34 ÷ 39.
liberstina [14]

Answer:

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8 0
3 years ago
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Express answer in exact form. Show all work for full credit.
Gennadij [26K]

Answer:

Area segment = 3/2 π - (9/4)√3 units²

Step-by-step explanation:

∵ The hexagon is regular, then it is formed by 6 equilateral Δ

∵ Area segment = area sector - area Δ

∵ Area sector = (Ф/360) × πr²

∵ Ф = 60° ⇒ central angle of the sector

∵ r = 3

∴ Area sector = (60/360) × (3)² × π = 3/2 π

∵ Area equilateral Δ = 1/4 s²√3

∵ The length of the side of the Δ = 3

∴ Area Δ = 1/4 × (3)² √3 = (9/4)√3

∴ Area segment = 3/2 π - (9/4)√3 units²

4 0
3 years ago
Show that if f(n) is O(g(n)) and d(n) is O(h(n)) then f(n) + d(n) is O(g(n) + h(n)).
GaryK [48]
f(n)\in\mathcal O(g(n)) is to say

|f(n)|\le M_1|g(n)|

for all n beyond some fixed n_1.

Similarly, d(n)\in\mathcal O(h(n)) is to say

|d(n)|\le M_2|h(n)|

for all n\ge n_2.

From this we can gather that

|f(n)+d(n)|\le|f(n)|+|d(n)|\le M_1|g(n)|+M_2|h(n)|\le M(|g(n)|+|h(n)|)

where M is the larger of the two values M_1 and M_2, or M=\max\{M_1,M_2\}. Then the last term is bounded above by

M(|g(n)|+|h(n)|)\le2M\max\{|g(n)|,|h(n)|\}

from which it follows that

f(n)+d(n)\in\mathcal O(\max\{g(n),h(n)\})
3 0
3 years ago
I dont understand this problem
Fiesta28 [93]

Answer:

D and F

Step-by-step explanation:

x^{2}+4x+4=12 can be rewritten as x^{2} +4x-8=0.

From there you can use the quadratic formula x=\frac{-b±\sqrt{b^{2}+4ac } }{2a} (ignore the A, I can‘t seem to remove it), where a=1, b=4, and c=-8.

x=\frac{-4±\sqrt{4^{2}-4(1)(-8)} }{2(1)}

Then you get x=\frac{-4±\sqrt{16+32}}{2}

Then you get x=\frac{-4±\sqrt{48} }{2}

So x=-2±\frac{4\sqrt{3} }{2}

Which means x=-2+2\sqrt{3} or -2-2\sqrt{3} which are choices D and F.

8 0
2 years ago
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stepladder [879]

Answer:

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Please brainlist!

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3 years ago
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