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3 years ago

Y=1/x-2+1 the domain and range of this function

Mathematics

1 answer:

Alex777 [14]3 years ago

8 0

Answer:

Domain=(-infinity,0)

Range: (-infinity,-1)

Step-by-step explanation:

Find the domain by finding where the equation is defined. The range is the set of values that correspond with the domain.

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How to find the perimeter of the triangle

Sliva [168]

Answer:

A. 26.2

Step-by-step explanation:

To find the perimeter of the triangle, you have to find the distances of all three lines and add them up.

Line AB

Let's start off by finding the distance of line AB.

We will use the formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point B is (4,-3).

To substitute the values, it will get to $d = \sqrt{(4--2)^{2}+(-3-5)^{2}}$ which in other words is $d = \sqrt{(4+2)^{2}+(-3-5)^{2}}$ .

Now we have to solve the parenthesis to get $d = \sqrt{(6)^{2}+(-8)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{36+64}$ .

Now we have to simplify the square root to $d = \sqrt{100}$ . In other words, that is $d = 10$ .

Line AB = 10

Line BC

Now let's find the distance of line BC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point B is (4,-3) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0-4)^{2}+(-6--3)^{2}}$ which in other words is $d = \sqrt{(0-4)^{2}+(-6+3)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(-4)^{2}+(-3)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{16+9}$ .

Now we have to simplify the square root to $d = \sqrt{25}$ . In other words, that is $d = 5$ .

Line BC = 5.

Line AC

Now let's fine the distance of line AC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0--2)^{2}+(-6-5)^{2}}$ which in other words is $d = \sqrt{(0+2)^{2}+(-6-5)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(2)^{2}+(-11)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{4+121}$ .

Now we have to simplify the square root to $d = \sqrt{125}$ . In other words, that is $d = 5\sqrt{5}$ . To round that, $d = 11.2$ .

Line AC = 11.2.

Perimeter of Triangle ABC

Perimeter of Triangle ABC = Line AB + Line BC + Line AC.

Perimeter of Triangle ABC = 10 + 5 + 11.2

Perimeter of Triangle ABC = 26.2

Hope this helped! If not, please let me know <3

3 0

3 years ago

7. We learned that in a plane, two lines perpendicular to the same line are parallel. Use the rectangular solid below to explain

Illusion [34]

Answer:

Step-by-step explanation:

6 0

4 years ago

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Complete the square to make a perfect square trinomial. Write the result as a binomial square. q2−23q

tia_tia [17]

Answer:

$\left(q-\dfrac{23}{2}\right)^2$

Step-by-step explanation:

Given the expression: $q^2-23q$

To complete the square, we follow these steps:

Step 1: Identify the coefficient of q

Coefficient of q=-23

Step 2: Divide the coefficient of q by 2

$=-\dfrac{23}{2}$

Step 3: Square your result from step 2 and add it to the equation

This gives us: $q^2-23q+\left(-\dfrac{23}{2}\right)^2$

We have now completed the square.

Step 4: Write the result as a binomial square.

To write it as a binomial square, pick the variable and add the term in the bracket.

Therefore:

$q^2-23q+\left(-\dfrac{23}{2}\right)^2=\left(q-\dfrac{23}{2}\right)^2$

5 0

3 years ago

Help please, I will give a brainly to whoever has it done first

Reil [10]

Answer:

297/40 or 7.425

Step-by-step explanation:

Hope this helped

To find the volume you got to multiply W by H and L

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3 years ago

What is the value of 7 in 0.750

GenaCL600 [577]

It is in the tenths place, 7/10

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3 years ago

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Y=1/x-2+1 the domain and range of this function​

Y=1/x-2+1 the domain and range of this function