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sp2606 [1]
3 years ago
6

A particular model of walkie-talkie can broadcast in a circular area. The radius of the broadcast area is 10,000 feet.

Mathematics
1 answer:
Nikolay [14]3 years ago
6 0
I think thatv  the answer is  a 314,000,000 ft ^2
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Hello <br> Can you help please find the answer c ?
lawyer [7]

Answer:

150 i think

Step-by-step explanation:

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7 0
3 years ago
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Simplify x 1/3(x1/2+2x²​
tigry1 [53]

Answer:

x^{\frac{5}{6} } +2x^\frac{7}{3}

Step-by-step explanation:

1.  x^1/3(x^1/2) = x^{\frac{5}{6} }

2. x^1/3(2x^2) = 2x^\frac{7}{3}

3. = x^{\frac{5}{6} } + 2x^\frac{7}{3}

4 0
3 years ago
1.What is the area of this parallelogram?
dexar [7]

Question 1: Option D

Area of the parallelogram = 132 cm²

Question 2: Option B

Area of the parallelogram = 3.60 ft²

Solution:

Question 1:

Base of the parallelogram = 6 + 5 = 11 cm

Height of the parallelogram = 12 cm

Area of the parallelogram = Base × Height

                                           = 11 × 12

Area of the parallelogram = 132 cm²

Option D is the correct answer.

Question 2:

Base of the parallelogram = 1.9 + 0.5 = 2.4 ft

Height of the parallelogram = 1.5 ft

Area of the parallelogram = Base × Height

                                           = 2.4 × 1.5

Area of the parallelogram = 3.60 ft²

Option B is the correct answer.

7 0
3 years ago
Please Solve this, it would be extremely helpful for me.
Sidana [21]

Answer:

Step-by-step explanation:

1) ΔCPD & ΔEPF

∠CPD = ∠EPF   { Vertically opposite angles}

∠CDP = ∠PFE {CD║EF, FD is transversal, Alternate interior angles are equal}

ΔCPD ≈ΔEPF  {AA criteria for similarity }

\frac{DC}{EF} =\frac{PC}{EP}\\\\\\\frac{27}{EF}=\frac{15}{7.5}\\\\

Cross multiply

EF * 15 = 27 * 7.5

EF =\frac{27*7.5}{15}\\\\

EF = 27 * 0.5

EF = 13.5 cm

ii) EF // AB, so Triangles ACB & ECF are similar triangles

\frac{AB}{EF}=\frac{AC}{EC}\\\\\frac{22.5}{13.5}=\frac{AC}{22.5}

AC= \frac{22.5*22.5}{13.5}\\\\AC=37.5 cm

AC = 37.5 cm

6 0
3 years ago
The monthly sales (in thousands of units) of a seasonal product are approximated by
Elodia [21]
 <span>Don't forget S is measured in thousands of units so you are solving for : 

100 < 74.5 + 43.75Sin(πt/6) 
25.5 < 43.75Sin(πt/6) 
Sin(πt/6) >25.5/43.75 = 0.582857 
ASrcSin(πt/6) > 0.62224 radians 
πt/6 > 0.62224 
t > 6 x 0.62224/π = 1.1884 (4dp) 

This initial value occurs when the sine value is increasing and it will reach its maximum value of 1 when Sin(πt/6) = Sinπ/2, that is when t = 3. 
Consequently, monthly sales exceed 100,000 during the period between t = 1.1884 and 4.8116 
[3 - 1.1884 = 1.8116 so the other extreme occurs at 3 + 1.8116] 

Note : on the basis of these calculations, January is 0 ≤ t < 1 : February is 1 ≤ t < 2 :....May is 4 ≤ t < 5 
So the period when sales exceed 100,000 occurs between Feb 6 and May 25 and annually thereafter.</span>
3 0
3 years ago
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