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Whitepunk [10]
3 years ago
10

How many different ways can the letters in the word "objects" be arranged?

Mathematics
1 answer:
sdas [7]3 years ago
6 0
It is known that 6 things can get arranged in 720 ways.
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Heeeeeeeeeeeeeeeeeelllllllllllllllllllllllppppppppppppppppppppppp
Dvinal [7]

Answer:

a) 50yd2

b) 7ft2

Step-by-step explanation:

a) 1/2 x b x b

1/2 x 4 x 25 = 50

b) 1/2 x b x h

1/2 x 4 x 3.5 = 7

3 0
3 years ago
23=x-9;x+14 I need help with this problem.
frez [133]
 is  this answer 
x=1 ;) 
6 0
3 years ago
A string of lights contains three lights. the lights are wired in series, so that if any light fails the whole string will go da
vazorg [7]
Let p_i=0.1 be a  probability that i-th light fails, where i=1,2,3. Then, q_i=1-p_i=1-0.1=0.9 is the probability that i-th light remain bright.
<span>If the lights fail independently of each other,  the probability that a string of lights will remain bright for two years is:
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q_1q_2q_3=(0.9)^3=0.729.





3 0
3 years ago
Write a quadratic function f whose zeros are 6 ad 5
dolphi86 [110]
46!! Stay safe! That’s the answe
3 0
3 years ago
1. Bob tried to answer the following question by finding the missing angle and rounding the answer to the nearest degree.
iren2701 [21]

Answer:

Part 1)

Bob's mistake was to have used the cosine instead of the sine

The measure of the missing angle is 53.13\°

Part 2) The surface area of the pyramid is 288\ cm^{2}

Step-by-step explanation:

Part 1)

Let

x----> the missing angle

we know that

In the right triangle o the figure

The sine of angle x is equal to divide the opposite side angle x to the hypotenuse of the right triangle

sin(x)=\frac{16}{20}

x=arcsin(\frac{16}{20})=53.13\°

Bob's mistake was to have used the cosine instead of the sine

Part 2) we know that

The surface area of the square pyramid is equal to the area of the square base plus the area of its four lateral triangular faces

so

SA=b^{2}+4[\frac{1}{2}(b)(h)]

where

b is the length side of the square

h is the height of the triangular lateral face

In this problem

h=b/2 -------> by an 45° angle

so

b=2h

sin(45\°)=\frac{h}{6\sqrt{2}}

h=6\sqrt{2}(sin(45\°))=6\ cm

Find the value of b

b=2(6)=12\ cm

Find the surface area

SA=12^{2}+4[\frac{1}{2}(12)(6)]=288\ cm^{2}

3 0
4 years ago
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