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SIZIF [17.4K]
3 years ago
11

1.Un número complejo multiplicado por su conjugado da siempre:

Mathematics
2 answers:
algol133 years ago
8 0
B. 2+5i dis the answer
Ivan3 years ago
7 0
1. C
2. D
3. A
4. D
5. B
6. B
7. B
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how for is Blaine's house if you must drive 6.5 hours through the woods at an average of 45mph to get there?
omeli [17]
Six and a half hours at a speed of 45 mph. 45 mph is forty five miles per hour. since there are six full hours, multiply six (time) by speed (45) to get the distance within the six full hours, which is 270 miles. you also have an additional half hour at a 45 mph speed, so rather than going 45 miles in one hour, you for half an hour, meaning you go 22.5 miles. add these two together (270 + 22.5) to get 292.5 miles. a faster way to get to this would be to multiply 6.5 * 45. hole this helps!
3 0
3 years ago
A parabola with the vertex at (1.5, 12.5) inscribes a segment of length 5 on x-axis. What is the area of the triangle formed by
damaskus [11]

Answer:

  20 units²

Step-by-step explanation:

The x-intercepts are symmetrically located around the x-coordinate of the vertex, so are at

  1.5 ± 5/2 = {-1, 4}

Using one of these we can find the unknown parameter "a" in the parabola's equation (in vertex form) ...

  0 = a(4 -1.5)² +12.5

  0 = 6.25a +12.5 . . . . . simplify

  0 = a +2 . . . . . . . . . . . divide by 6.25

  -2 = a

Then the standard-form equation of the parabola is ...

  y = -2(x -1.5)² +12.5 = -2(x² -3x +2.25) +12.5

  y = -2x² +6x +8

This tells us the y-intercept is 8. Then the relevant triangle has a base of 5 units and a height of 8. Its area is given by the formula ...

  A = (1/2)bh = (1/2)(5)(8) = 20 . . . . units²

5 0
3 years ago
Cobalt-56 has a decay constant of 8.77 × 10-3 (which is equivalent to a half life of 79 days). How many days will it take for a
sweet-ann [11.9K]
As I read this one, is just a decay exponential equation... so

A = P(1 + r)ᵗ   where "t" is days passed. . hmm in this case is decay, so negative rate  A = P(1 - r)ᵗ, and the decimal amount would be 0.00877 for the rate

62% of P, the original value, is just 0.62P, now... if we hmm take P as just 1, it could be any amount, but 62% of 1,000,000 is just 62% of 1 times 1,000,000

so, for the sake of comparing it with a percentage, 1 will do

\bf 0.62P=P(1-0.00877)^t\implies P=1\implies 0.62=(1-0.00877)^t
\\\\\\
ln(0.62)=ln[(1-0.00877)^t]\implies ln(0.62)=t\cdot  ln(1-0.00877)
\\\\\\
\cfrac{ln(0.62)}{ln(0.99123)}=t
7 0
3 years ago
1. what is the area of the board shown on the scale drawing? explain how you found the area.2. how can Adam use the scale factor
bixtya [17]

Answer:

1. 1800 square cm.

2. See below

3. 45000 square cm.

Explanation:

Part 1

The dimensions of the drawing are 36cm by 50cm.

\begin{gathered} \text{The area of the board}=36\times50 \\ =1800\operatorname{cm}^2 \end{gathered}

Part 2

Given a scale factor, k

If the area of the scale drawing is A; then we can find the area of the actual board by multiplying the area of the scale drawing by the square of k.

Part 3

\begin{gathered} \text{Area of the scale drawing}=1800\operatorname{cm}^2 \\ \text{Scale Factor,k=5} \end{gathered}

Therefore, the area of the actual drawing will be:

\begin{gathered} 1800\times5^2 \\ =45,000\operatorname{cm}^2 \end{gathered}

7 0
1 year ago
Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime i
OLEGan [10]
<h2>Answer with explanation:</h2>

Let \mu be the population mean lifetime of circulated $1 bills.

By considering the given information , we have :-

H_0:\mu=18\\\\H_a:\mu\neq18

Since the alternative hypotheses is two tailed so the test is a two tailed test.

We assume that the lifetime of circulated $1 bills is normally distributed.

Given : Sample size :  n=50 , which is greater than 30 .

It means the sample is large so we use z-test.

Sample mean : \overline{x}=18.8

Standard deviation : \sigma=2.8

Test statistic for population mean :-

z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02

The p-value= 2P(z>2.02)=0.0433834

Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.

6 0
3 years ago
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