Select the sequence of transformations that will carry triangle A onto triangle A'.

8 + 3x=29 +2x I need answer ASAP

borishaifa [10]

Answer:

x = 21

Step-by-step explanation:

8 + 3x = 29 + 2x

8 + 3x - 2x = 29

8 + x = 29

x = 29 - 8 = 21

x = 21

3 0

2 years ago

Solve the system using substitution x+5y=0 3y+2x=-21

Crazy boy [7]

Answer:

$\left \{ {{x = -15} \atop {y=3}} \right.$

Step-by-step explanation:

$\left \{ {{x + 5y = 0} (1) \atop {3y + 2x = -21}(2)} \right. \left \{ {{x = -5y} \atop {3y + 2x = -21}} \right. \left \{ {{x = -5y} \atop {3y + 2*(-5y) = -21}} \right. \left \{ {{x = -5y} \atop {3y - 10y = -21}} \right. \\ \left \{ {{x = -5y } \atop {-7y = -21}} \right. \left \{ {{x = -5y} \atop {y= 3 }} \right. \left \{ {{x = -5 * 3} \atop {y=3}} \right.=> \left \{ {{x=-15} \atop {y=3}} \right.$

5 0

2 years ago

Read 2 more answers

The set of ordered pairs represents a

Murrr4er [49]

Answer:

The rule that represents the function is $y=x^2+1$ therefore the function is $f(x)=x^2+1$

Step-by-step explanation:

We have 5 ordered pairs in the plane xy. This means that every pair has the form (x, y).

Then, we have 5 values of x, which will give us 5 values of y, using the rule that represents the function.

The easy evaluation is that when x=0, the value of y is y=1, and then we can evaluate the rule for x=-1, and x=1, the value of y is the same, y=2. We can see here that we have a parabolic function, that is not centered in the origin of coordinates because when x=0, y=1.

So we propose the rule $y=x^2+1$ which is correct for the first 3 values of x.

Now, we evaluate the proposed rule when x=2, and when x=3. This evaluations can be written as

$f(2)=2^2+1=5$

$f(3)=3^2+1=10$

Therefore, the rule is correct, and the function is

$f(x)=x^2+1$

8 0

3 years ago

What would the total area be??

timurjin [86]

Answer:

1,560,000 mm²

Step-by-step explanation:

A= $\frac{(1600*600)}{2}$ This would make shape A's area 480,000.

B= $\frac{(600*600)}{2}$ This would make shape B's area 180,000.

C= $\frac{(1000*600)}{2}$ This would make shape C's area 300,000.

D= 1000×600 This wouls make D's area 600,000.

Now you add up all of the areas;

480,000+180,000+300,000+600,000= 1,560,000 mm²

6 0

3 years ago

Read 2 more answers

An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of

ANEK [815]

Answer:

(a) 0.152

(b) 0.789

(c) 0.906

Step-by-step explanation:

Let's denote the events as follows:

F = The word free occurs in an email

S = The email is spam

V = The email is valid.

The information provided to us are:

The probability of the word free occurring in a spam message is, $P(F|S)=0.60$
The probability of the word free occurring in a valid message is, $P(F|V)=0.04$
The probability of spam messages is,

$P(S)=0.20$

First let's compute the probability of valid messages:

$P (V) = 1 - P(S)\\=1-0.20\\=0.80$

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

$P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The probability that a message contains the word free is:

$P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\$

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is spam provided that it contains free is:

$P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\$

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is valid provided that it does not contain free is:

$P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566$

The probability that a message is valid provided that it does not contain free is approximately 0.906.

4 0

3 years ago