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olga nikolaevna [1]
3 years ago
10

Evaluate the discriminant. How many real and imaginary solutions does each have?

Mathematics
1 answer:
Sav [38]3 years ago
4 0
Evaluate the discriminant. How many real and imaginary solutions does each have?

3x^2 - x + 3 = 0

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What is another way to write 49*12? 1. 50*10<br> 2. 50*10+2<br> 3. 50*12-12 <br> 4. 50+12-12
Umnica [9.8K]
49×12= 588
1. 50×10=500
2. 50×10+2= 502 
3. 50×12-12= 588
4. 50+12-12=50

Another way to write 49×12 is 50×12-12

4 0
3 years ago
24 POINTS!!!!!!
nika2105 [10]
5x - 2 equivalent to to  2x−2+3x

answer is B. 2x−2+3x
7 0
3 years ago
We are studying the proportion of cities that have private garbage collector. We want a maximum error of 0.10 and a 95% confiden
True [87]

Answer:

The minimum number of cities we need to contact is 96.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence interval 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

Z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence interval

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

In this problem, we have that:

p = 0.5, M = 0.1

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.1 = 1.96*\sqrt{\frac{0.5*0.5}{n}}

0.1\sqrt{n} = 0.98

\sqrt{n} = 9.8

(\sqrt{n})^{2} = (9.8)^{2}

n = 96

The minimum number of cities we need to contact is 96.

8 0
3 years ago
Andre is running in an 80-meter hurdle race. There are 8 equally-space hurdles on the race track. The first hurdle is 12 meters
evablogger [386]

Answer:

600 and 3

Step-by-step explanation:

6 0
3 years ago
Help no time plzzzz plz plz​
Sav [38]

Answer:

FALSE

Step-by-step explanation:

<E in ∆AED ≅ <E in ∆CEB.

Both are 90°.

Side ED ≅ Side EB

Side AD ≅ Side CB.

Thus, two sides (ED and AD) and a non-included angle (<E) of ∆AED are congruent to corresponding two sides (EB and CB) and a non-included angle (<E) of ∆CEB. Therefore, by A-S-S Congruence Theorem, both triangles are congruent to each other not by SSS.

8 0
3 years ago
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