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Vinil7 [7]
3 years ago
10

Opposite angles of an inscribed quadrilateral are supplementary?

Mathematics
2 answers:
netineya [11]3 years ago
8 0

Answer:

An inscribed quadrilateral is any four sided figure whose vertices all lie on a circle. ... This conjecture give a relation between the opposite angles of such a quadrilateral. It says that these opposite angles are in fact supplements for each other. In other words, the sum of their measures is 180 degrees.

tangare [24]3 years ago
4 0

Answer:

yes

Step-by-step explanation:

according to cyclic quadrilateral properties the oppoaite angles are supplementry

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Solve x2 + 2x + 9 = 0
Keith_Richards [23]
x^2+2x+9=0 \\ \\
a=1 \\ b=2 \\ c=9 \\ b^2-4ac=2^2-4 \times 1 \times 9=4-36=-32 \\ \\
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-2 \pm \sqrt{-32}}{2 \times 1}=\frac{-2 \pm \sqrt{-16 \times 2}}{2}=\frac{-2 \pm 4i\sqrt{2}}{2}=-1 \pm 2i\sqrt{2}

The answer is D.
8 0
3 years ago
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What is the total number of the side on 8 triangle
Lelu [443]
Total number of triangle is 632
With 3 diagonal endpoints are 56
5 0
3 years ago
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A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i
garri49 [273]

Answer:

The degrees of freedom first given by:  

df=n-1=12-1=11  

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

t_{\alpha}= 1.796

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :  

Null hypothesis:\mu \leq 5  

Alternative hypothesis:\mu > 5  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

The degrees of freedom first given by:  

df=n-1=12-1=11  

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

t_{\alpha}= 1.796

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0
3 years ago
You have 42,784 grams of a radioactive kind of curium. If its half-life is 18 years, how much will be left after 72 years?
s344n2d4d5 [400]

Answer:

2,674.14 g

Step-by-step explanation:

Recall that the formula for radioactive decay is

N = N₀ e^(-λt)

where,

N is the amount left at time t

N₀ is the initial amount when t=0, (given as 42,784 g)

λ = coefficient of radioactive decay

  = 0.693 ÷ Half Life

  = 0.693 ÷ 18

  = 0.0385

t = time elapsed (given as 72 years)

e = exponential constant ( approx 2.7183)

If we substitute these into our equation:

N = N₀ e^(-λt)

= (42,787) (2.7183)^[(-0.0385)(72)]

= (42,787) (2.7183)^(-2.7726)

=  (42,787) (0.0625)

= 2,674.14 g

6 0
3 years ago
What is the common ratio of the geometric sequence: -1/12.-1/2,-3,18,...
Minchanka [31]
\frac{1}{6} : Common ratio

Check:
=-\frac{1}{12} ÷ \frac{1}{6}
=-\frac{1}{12} × 6
=-\frac{1}{2}

=-\frac{1}{2} ÷ \frac{1}{6}
=- \frac{1}{2} × 6
=-3
8 0
3 years ago
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