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Inga [223]
3 years ago
14

A net of a square pyramid is shown. The total area of the pyramid's triangular faces is 80 cm'. What is the area of the pyramid'

s square base in centimeters squared?​

Mathematics
1 answer:
ankoles [38]3 years ago
5 0
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33, 25, 42, 25, 31, 37, 46,29,38<br>What is the first quartile of the data?​
NISA [10]
answer:
Step by step solution:
1) put the numbers in order:
25,25,29,31,33,37,38,42,46
2) find the median:
25,25,29,31,33,37,38,42,46
33
3) find the median of the first quartile:
25,25,29,31
25 and 29
25+29= 54
54/2
27
Q1= 27

6 0
3 years ago
BC =<br> Round your answer to the nearest hundredth.<br> ?<br> C<br> B<br> 50°<br> 7<br> А
umka21 [38]

Answer:

The answer is 8.34

U have to use the SOH CAH TOA method

In this case I used tan

so,

tan 50= BC/7

solve for BC= 7* tan 50= 8.34

6 0
3 years ago
Evaluate f(x) =3x+8 for x=1
SpyIntel [72]
X= 11
If you plug in 1 for X in the equation, you will get:
 f(x) =3+8
3+8=11
Hope this helps.

5 0
4 years ago
Match the equations with their solutions over the interval [0, 2π].
nignag [31]

I solved this using a scientific calculator and in radians mode since the given x's is between 0 to 2π. After substitution, the correct pairs are:

cos(x)tan(x) – ½ = 0 → π/6 and 5π/6

cos(π/6)tan(π/6) – ½ = 0

cos(5π/6)tan(5π/6) – ½ = 0

 

sec(x)cot(x) + 2 = 0 → 7π/6 and 11π/6

sec(7π/6)cot(7π/6) + 2 = 0

sec(11π/6)cot(11π/6) + 2 = 0

 

sin(x)cot(x) + 1/sqrt2 = 0 → 3π/4 and 5π/4

sin(3π/4)cot(3π/4) + 1/sqrt2 = 0

sin(5π/4)cot(5π/4) + 1/sqrt2 = 0

 

csc(x)tan(x) – 2 = 0 → π/3 and 5π/3

csc(π/3)tan(π/3) – 2 = 0

csc(5π/3)tan(5π/3) – 2 = 0

5 0
4 years ago
Read 2 more answers
Please help me ASAP!!! I will give 20 points..... bad answers will be reported
belka [17]
150? I think thats the anwser
3 0
3 years ago
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