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Semmy [17]
2 years ago
9

If x = (7+4√3), then the value of x^2 + 1/x^2 is​

Mathematics
1 answer:
Blizzard [7]2 years ago
5 0

Answer:

194.

Step-by-step explanation:

x = (7+4√3)

x^2 =  x = (7+4√3)^2

= 49 + 48 + 56√3

= 97 + 56√3

x^2 + 1/x^2 =    97 + 56√3 + 1/(97 + 56√3)

= 194.

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What’s the answer to this?
inna [77]

Since the triangle is equilateral, all its angles are equal to 60°

AO is the bisector⇒∠OAD = 30°

AO is the hypotenuse, ∠OAD = 30°⇒

OD=5*2=10m

By the Pythagorean theorem

10^2=AD^2+5^2\\AD=\sqrt{100-25} \\AD=\sqrt{75}

AC=AD+DC=2\sqrt{75}=10\sqrt{3}

S=\frac{a^2\sqrt{3} }{4}

S=\frac{(10\sqrt{3})^2\sqrt{3}  }{4} =\frac{300\sqrt{3} }{4} =75\sqrt{3} m^2

Answer: A) S=75\sqrt{3}m²

P.S. Hello from Russia and sorry for my bad english :^)

3 0
3 years ago
Round 17,519 to the underlined place value.<br><br> 17,500<br> 20,000<br> 18,000<br> 17,000
Ira Lisetskai [31]
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5 0
3 years ago
When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected
jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

N=\frac{12.36}{0.03+0.55^t}

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

N=\frac{12.36}{0.33+0.55^0}

N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12

A. Initially, there were 12 deer.

B. N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

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Answer:

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