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3 years ago

How many solutions does this graph have?

Mathematics

1 answer:

Aloiza [94]3 years ago

3 0

Answer:

no solutions

Step-by-step explanation:

The graph never intersect.

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Segment FG begins at point F(-2, 4) and ends at point G (-2, -3). Segment FG is translated by (x, y) → (x – 3, y + 2) and then r

Mariana [72]

Answer:

The length of the segment F'G' is 7.

Step-by-step explanation:

From Linear Algebra we define reflection across the y-axis as follows:

$(x',y')=(-x, y)$ , $\forall\, x, y\in \mathbb{R}$ (Eq. 1)

In addition, we get this translation formula from the statement of the problem:

$(x',y') =(x-3,y+2)$ , $\forall \,x,y\in \mathbb{R}$ (Eq. 2)

Where:

$(x, y)$ - Original point, dimensionless.

$(x', y')$ - Transformed point, dimensionless.

If we know that $F(x,y) = (-2, 4)$ and $G(x,y) = (-2,-3)$ , then we proceed to make all needed operations:

Translation

$F''(x,y) = (-2-3,4+2)$

$F''(x,y) = (-5,6)$

$G''(x,y) = (-2-3,-3+2)$

$G''(x,y) = (-5,-1)$

Reflection

$F'(x,y) = (5, 6)$

$G'(x,y) = (5,-1)$

Lastly, we calculate the length of the segment F'G' by Pythagorean Theorem:

$F'G' = \sqrt{(5-5)^{2}+[(-1)-6]^{2}}$

$F'G' = 7$

The length of the segment F'G' is 7.

8 0

3 years ago

What is the pattern here?

tigry1 [53]

To get the number at the back, you would need to multiply the number in front by 3 and minus 1.
The answer would be A

8 0

3 years ago

50 PTS ANSWER ALL <3333333

11Alexandr11 [23.1K]

QUESTION 33

The length of the legs of the right triangle are given as,

6 centimeters and 8 centimeters.

The length of the hypotenuse can be found using the Pythagoras Theorem.

${h}^{2} = {6}^{2} + {8}^{2}$

${h}^{2} = 36+ 64$

${h}^{2} = 100$

$h = \sqrt{100}$

$h = 10cm$

Answer: C

QUESTION 34

The triangle has a hypotenuse of length, 55 inches and a leg of 33 inches.

The length of the other leg can be found using the Pythagoras Theorem,

${l}^{2} + {33}^{2} = {55}^{2}$

${l}^{2} = {55}^{2} - {33}^{2}$

${l}^{2} = 1936$

$l = \sqrt{1936}$

$l = 44cm$

Answer:B

QUESTION 35.

We want to find the distance between,

(2,-1) and (-1,3).

Recall the distance formula,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the values to get,

$d=\sqrt{( - 1-2)^2+(3- - 1)^2}$

$d=\sqrt{( - 3)^2+(4)^2}$

$d=\sqrt{9+16}$

$d=\sqrt{25}$

$d = 5$

Answer: 5 units.

QUESTION 36

We want to find the distance between,

(2,2) and (-3,-3).

We use the distance formula again,

$d=\sqrt{( - 3-2)^2+( - 3- 2)^2}$

$d=\sqrt{( - 5)^2+( - 5)^2}$

$d=\sqrt{25+25}$

$d=\sqrt{50}$

$d=5\sqrt{2}$

Answer: D

8 0

2 years ago

Read 2 more answers

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0

3 years ago

Last year, Austin opened an investment account with $5400. At the end of the year, the amount in the account had increased by

zlopas [31]

Answer:

5400 + 351 = 5751

Step-by-step explanation:

4 0

2 years ago