Find an equation for the line that passes through (-4, 8) and (3, -7). What is the slope? Where does the line intersect the x-axis and the y-axis?

Answers

Answer 1

Answer: the equation of the line is : 7y +15x = -4
the slope is m= -15/7
intersect with x at (-4/15 ,0)
intersect with y at (0, -4/7)

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What is 1÷0?and is it possible

Answers

Instead of just assigning one willy nilly, we say that infinity isn't a number, and that 1/0 is undefined. When something is divided by 0, why is the answer undefined? The reason is related to the associated multiplication question. If you divide 6 by 3 the answer is 2 because 2 times 3 IS 6.

Hi There! :)

What is 1÷0?and is it possible

Not Possible! :P

Segment CD is given with vertices C(a,b) and D(2a,b). Find the length of segment C'D', the image of segment CD after a dilation with a scale factor of 2.

Answers

the dilation is a scale factor of 2, that is:

C'D' = 2(CD)

the length of CD is given by:
CD = √( (a - 2a)^2 + (b - b)^2 )

CD = √(a^2) = a

hence the new segment is 2a long.

Explain: How could you use cubes to act out a problem

Answers

From breaking up the problems and sorting them out

to break the problem apart to make it easier to understand

Practice simplifying rational expressions with negative exponents.

Answers

Answer:

Part 1:- option first is correct

$-(1)/(2)ab^(12)$

Part 1:- option third is correct

$(w^(10))/(3y^(4))$

Step-by-step explanation:

Given:

The given ration expressions are.

1. $(-2a^(2)b^(4))/(4ab^(-8))$

2. $(-5w^(4)y^(-2))/(-15w^(-6)y^(2))$

We need to simplify the given expressions.

Solution:

Part 1:-

Given expression is

$=(-2a^(2)b^(4))/(4ab^(-8))$

Using law of exponents $(x^(m) )/(x^(n) ) = x^((m-n))$

$=-(a^(2-1)b^(4-(-8)))/(2)$

$=-(a^(1)b^(4+8))/(2)$

$=-(ab^(12))/(2)$

$=-(1)/(2)ab^(12)$

Therefore, option first $-(1)/(2)ab^(12)$ is correct.

Part 2:-

Given expression is.

$=(-5w^(4)y^(-2))/(-15w^(-6)y^(2))$

Using law of exponents $(x^(m) )/(x^(n) ) = x^((m-n))$

$=(w^(4-(-6))y^(-2-2))/(3)$

$=(w^(4+6)y^(-4))/(3)$

$=(w^(10)y^(-4))/(3)$

$=(w^(10))/(3y^(4))$

Therefore, third option $(w^(10))/(3y^(4))$ is correct.

Find the range 4.7 6.3 5.4 3.2 4.9

Answers

Answer:

3.1.

Step-by-step explanation:

We have been given a data set and we are asked to find the range of our given data set.

4.7, 6.3, 5.4, 3.2, 4.9

First of all let us arrange our data set in ascending order.

3.2, 4.7, 4.9, 5.4, 6.3

Since we know that range is the difference between the greatest value and lowest value of the data set.

$\text{Range}=\text{ The greatest value of data set - The lowest value of data set}$

We can see that 6.3 is the greatest data point and 3.2 is the lowest data point of our given data set.

Upon substituting these values in range formula we will get,

$\text{Range of our given data set}=6.3-3.2$

$\text{Range of our given data set}=3.1$

Therefore, the range of our given data set is 3.1.

Simple....

Range= Biggest number-Smallest Number

6.3-3.2=3.1

Thus, your answer.

1) Find the inverse function of f(x)=1/2x+32)Use composition to verify that they are inverse relations?
3) f^ Domain : Range:
4) f^-1 Domain : Range:

Answers

y = 1/2x + 3
change x and y

x = 1/2y + 3
now find the value of y and that is inverse function

x - 3 =1/2 y
y = 2x- 6

f-(x) = 2x - 6

for both domain : ( - ∞ , + ∞ )

range f (x) : ( - ∞ , + ∞ ) - { 0 }
range f-(x) :( -∞ , +∞)

To find the inverse function you need to change $f(x)$ (call it $y$ ) and $x$ , then solve for $y$ :

$y = (1)/(2)x+3 \n x = (1)/(2)y + 3 \n x - 3 = (1)/(2)y \n 2x-6 = y$

So now you have $f^(-1)(x) = 2x-6$ .

Composition to prove inverse relation: $f \circ f^(-1) (x) = x$ :

$f(f^(-1)(x)) = (1)/(2)(2x-6)+3 = x - 3 + 3 = x \square$

Domain and Range of both functions is Real numbers since they are both linear equations.

Other Questions

Just need help with question A

What is the point-slope form of an equation with the slope of 3/5 that passes through the point (10, –2)?

At 12 noon, thetemperature was3°F. Then thetemperature fellsteadily and reached-1°F at 2:00 PM.

Find the slope in the photo below: