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GREYUIT [131]
2 years ago
5

Compare positive and negative slope for linear relationships. How are they the same? How are they different?

Mathematics
1 answer:
Gekata [30.6K]2 years ago
5 0

Answer:

Please check explanations

Step-by-step explanation:

Slopes are generally the ratio of rise(vertical difference) to fall (horizontal difference) for set of points lying on a straight line

Positive slopes show a positive relationship between the two things we are considering while a negative one shows otherwise

They are the same in the sense that they actually represent a constant change between the coordinates of the points in the relationship we are considering.

They are different in the sense that while one represents a positive change, whereby both axes move in the same direction ( rise at same rate, fall at same rate) , the second represent a relationship in which they move in opposite directions. While one falls, the other rises and while the other rises, the second one falls which indicates an opposing movement type between the two

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Write 32/21 as a mixed number.
hammer [34]

Answer:

1 11/21

Step-by-step explanation:

In order to convert 32/21 to mixed number, you follow these steps:

Divide the numerator by the denominator

32 ÷ 21 = 1 with a remainder of 11

Write down the whole number 1 and then write down the remainder as new numerator (11) above the denominator (21), As below:

1  11/21

6 0
2 years ago
Read 2 more answers
Which of the following function types exhibit the end behavior f(x)-->0 as x --> -infinity?
ipn [44]
Let's consider the functions one by one.

i) y=x^n,

x --> -infinity means that x is a very very small negative number. To model it in our mind let's think of -10^{15}. This number to an even power becomes 10^{30}, 10^{60}... etc.

Indeed, we can see that the smaller the x, the greater the value x^n. In fact, as x --> -infinity,  f(x)-->+infinity.

ii)
y=x, 

this clearly means that the behaviors of x and y are identical.

As x --> -infinity, y --> -infinity as well.


iii) y=|x|,

We can think of x --> -infinity, again, as a very very small number, like -10^{15}. For this value of x, y is |-10^{15}|, that is 10^{15}.

Indeed, the smaller the x, the greater the y. As in the function in part (i), as x --> -infinity,  f(x)-->+infinity.

iv) 

y=1/x

consider the values of y for the following values of x:

for x=-10, -100, -1000, y is respectively -0.1, -0.01, -0.001.

We can see that the smaller the x, the closer y gets to 0.

Thus, <span>f(x)-->0 as x --> -infinity
</span>
v) n'th root of x in not defined for negative x, when n is even.

vi) y=b^x, b>0

Here note that b cannot be equal to 1, otherwise the function is not exponential.

Let b=5, consider the values of y for the following values of x:

for x=-10, -100, -1000, y is respectively \displaystyle{ \frac{1}{5^{10}},  \frac{1}{5^{100}}, \frac{1}{5^{1000}}.

That is, the smaller the value of x, the closer y gets to 0.

Thus, <span>f(x)-->0 as x --> -infinity</span>
5 0
3 years ago
Can Someone Please Help Me Out with My Math Homework!
yulyashka [42]
49a. r=9 & r=3
49b. r=59048 & r=19683
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8 0
3 years ago
Solve y3 = 125.<br><br> a.y = 5 <br> b.y = 25 <br> c.y = 11.2 <br> d. y = 15
Korolek [52]
Your answer is:

a. y=5
7 0
3 years ago
Read 2 more answers
Find the square. (1/4A + 1/4B)^2<br><br> a) 1/16A^2 + 1/8AB + 1/16B^2<br> b) 1/4A^2 + 1/8AB + 1/4B^2
alisha [4.7K]

(a + b)^2 = a^2 + 2ab + b^2

(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =

= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B^2

Answer: A.

Another way to solve by factoring 1/4.

(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =

= [\dfrac{1}{4}(A + B)]^2

= (\dfrac{1}{4})^2(A + B)^2

= \dfrac{1}{16}(A^2 + 2AB + B^2)

= \dfrac{1}{16}A^2 + \dfrac{1}{16}2AB + \dfrac{1}{16}B)^2

= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B)^2

6 0
3 years ago
Read 2 more answers
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