Write the equation in point-slope form that represents the graphed line.

What is the perimeter of the triangle? Round to the nearest tenth.

Cerrena [4.2K]

Answer: 19.8 Units

Step-by-step explanation:

The triangle is shown in the attached figure, and have three vertices: $AB$ , $AC$ and $BC$ .

Each vertice represents a point in the cartesian plane and the length of each side of the triangle is the modulus of each vector.

In this sense, the modulus $m$ is calculated as shown:

$m=\sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$

Hence, for each vector:

$AB=\sqrt{(-2-3)^{2} + (3-0)^{2}}=\sqrt{34}$

$AC=\sqrt{(-2-(4))^{2} + (3-(-3))^{2}}=\sqrt{40}$

$BC=\sqrt{(-4-3)^{2} + (-3-0)^{2}}=\sqrt{58}$

Now, the perimeter of a triangle is the sum of its three sides:

$AB+AC+BC=\sqrt{34} + \sqrt{40} + \sqrt{58}$

Finally:

$AB+AC+BC=19.77 \approx 19.8$

6 0

3 years ago

RS = 12, ST = 2x, RT = 34

Harrizon [31]

Answer:

x=11

Step-by-step explanation:

34(RT) minus 12(RS)=22. 22 divided by 2 equals 11.

6 0

3 years ago

Does the chart below represent a proportional relationship? What is the rate of change?

Lilit [14]

This chart represents a proportional relationship. The rate of change will be 4.

<h3>What is the constant of proportionality?</h3>

Firstly, there is a proportional relationship.

Two values x and y are said to be in a proportional relationship if x=ky, where x and y are variables and k is a constant.

The constant k is called the constant of proportionality.

Given;

x y

1 4

2 8

3 12

4 16

Here, x = ky

put the first value

k = 1/4

This chart represents a proportional relationship.

So the rate of change will be

$Rate = \dfrac{y_2 - y_1}{x_2 - x_1}\\\\Rate = \dfrac{8 - 4}{2 - 1}\\\\Rate = 4$

Learn more about proportional ;

brainly.com/question/20048815

#SPJ1

5 0

2 years ago

Please help me with these problems

S_A_V [24]

Answer:

1.a=2

2. C x=2 and x=-3

Step-by-step explanation:

The standard form for the quadratic function is

ax^2 +bx+c

so we need to rewrite the function to be in this form

2x^2 -10 = 7x

Subtract 7x from each side

2x^2 -7x-10 = 7x-7x

2x^2 -7x-10 = 0

a =2, b= -7 c=-10

2. The quadratic formula is

-b ± sqrt(b^2 -4ac)

----------------------------

2a

2x^2 + 2x=12

Lest get the equation in proper form

2x^2 + 2x-12 = 12-12

2x^2 +2x-12 =0

a=2 b=2 c=-12

Lets substitute what we know

-2 ± sqrt(2^2 -4(2)(-12))

----------------------------

2(2)

-2 ± sqrt(4+96)

----------------------------

2(2)

-2 ± sqrt(100)

----------------------------

4

-2 ± 10

----------------------------

4

-2 + 10 -2-10

----------- and --------------

4 4

8/4 and -12/4

2 and -3

8 0

3 years ago

Refer to the Trowbridge Manufacturing example in Problem 2-35. The quality control inspection proce- dure is to select 6 items,

Ivanshal [37]

Answer:

77.64% probability that there will be 0 or 1 defects in a sample of 6.

Step-by-step explanation:

For each item, there are only two possible outcomes. Either it is defective, or it is not. The probability of an item being defective is independent of other items. 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.

The true proportion of defects is 0.15

This means that $p = 0.15$

Sample of 6:

This means that $n = 6$

What is the probability that there will be 0 or 1 defects in a sample of 6?

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = C_{6,0}.(0.15)^{0}.(0.85)^{6} = 0.3771$

$P(X = 1) = C_{6,1}.(0.15)^{1}.(0.85)^{5} = 0.3993$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3771 + 0.3993 = 0.7764$

77.64% probability that there will be 0 or 1 defects in a sample of 6.

5 0

3 years ago