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zzz [600]
3 years ago
10

Given only a compass and straightedge, Greeks were able to construct only

Mathematics
1 answer:
OleMash [197]3 years ago
7 0
<h3>Answer: False</h3>

A regular heptagon (7-sided figure) cannot be constructed with a compass and straightedge.

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Someone help and please make sure the answer is right :)
Goryan [66]
I got x= -7. hope that helps you out!
8 0
3 years ago
together,sarah's two cats have the same mass as her dog,11 kg.the mass of one cat is 1500 g greater than the mass of the other.
Alika [10]
Let (a) be the first cat
let (b) be the second cat
a +b=11kg-----(1)
a=b+1500g⇒(1.5 kg)-------(2)
so now u have a system of two unknowns
a=11-b
substitute a in (2)
11-b=b+1.5
11-1.5=2b
2b=9.5
b=9.5/2=4.75
subs. b in (1)
a+b=11
a+4.75=11
a=11-4.75=6.25
so cat (a) is 6.25 kg
and (b) is 4.75 kg
hope I helped



6 0
3 years ago
24!!!!!!!!!!!!!!!!!!!!!
sashaice [31]
The two triangles on the side are 18 in^2 and the square in the middle is 36 in^2.  So in total, it has an area of 72 in^2
8 0
3 years ago
Suppose the force acting on a column that helps to support a building is a normally distributed random variable X with mean valu
Sauron [17]

Answer:

(1) 0.4207

(2) 0.7799

Step-by-step explanation:

Given,

Mean value,

\mu = 15.0

Standard deviation,

\sigma = 1.25

(1) P(X ≥ 17.5) = 1 - P( X ≤ 17.5)

= 1- P(\frac{x-\mu}{\sigma} \leq \frac{17.5-\mu}{\sigma})

=1-P(z\leq \frac{17.5 - 15}{1.25})

=1-P(z\leq \frac{2.5}{1.25})

=1-P(z\leq 2)

=1- 0.5793   ( By using z-score table )

= 0.4207

(2) P(14 ≤ X ≤ 18) = P(X ≤ 18) - P(X ≤ 14)

=P(z\leq \frac{18 - 15}{1.25}) - P(z\leq \frac{14 - 15}{1.25})

=P(z\leq \frac{3}{1.25}) - P(z\leq -\frac{1}{1.25})

=P(z\leq 2.4) - P(z\leq -0.8)

= 0.9918 - 0.2119

= 0.7799

8 0
3 years ago
Find the missing lengths of the sides.
velikii [3]
The first one pretty sure :))))
4 0
3 years ago
Read 2 more answers
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