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Damm [24]
2 years ago
5

Negative fractions on the number line

Mathematics
1 answer:
Elodia [21]2 years ago
3 0

Answer:

-1.2

Step-by-step explanation:

each mark is .20 so yeah

Pls mark brainliest!

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Mr. Fitch hires a landscape architect to renovate his front yard the rectangular yard is 30 feet long is the area of the front y
Vinil7 [7]

Answer:

The width of the yard is 20 ft

Step-by-step explanation:

In this particular question, we are asked to calculate the width of the rectangular front yard given the area of the front yard and the length of the front yard.

Mathematically, the area of the front yard is calculated by multiplying the length of the front yard by the breadth of the front yard.

Let’s say A = L * W

Substituting the known values, the unknown width would thus be:

W = A/L = 600 sq.ft / 30ft

W = 20 ft

7 0
3 years ago
If AD is the altitude of BC, what is the slope of AD?​
koban [17]

Answer:

D

Step-by-step explanation:

First, find the slope of BC :

⇒ m (BC) = -1 - 5 / -3 - 5

⇒ m (BC) = -6/-8

⇒ m (BC) = <u>3/4</u>

Now, since AD is the altitude of BC, we can say the altitude is perpendicular to BC.

Hence, the slope of a perpendicular line is equal to the negative reciprocal of the original line.

⇒ m (AD) = - (1/m(BC))

⇒ m (AD) = - (1/(3/4))

⇒ m (AD) = -4/3

4 0
2 years ago
Justify,All functions are relations,but not all relations are functions
Lana71 [14]

Answer: yes this statement is true All functions are relations,but not all relations are functions.

Step-by-step explanation:

Its true because if you think of it like this

a function is a output and like the inner circle

and relations is the outer circle.

Hope this helps :)

4 0
3 years ago
What are the intercept(s) of ƒ(x) = x2?
uranmaximum [27]

There is only one intercept at the origin - coordinates (0, 0).

5 0
3 years ago
Read 2 more answers
Can the mean value theorum be applied to the function f(x)=1/x^2 on the interval [-2, 1]? Explain.
Temka [501]

Answer:

No, it can not be applied.

Step-by-step explanation:

f(x) = 1/x²

f(x) is a polynomial that is not continuous

As,

f(x) = 1/0 is undefines

Secondly, it is not differentiable (i.e. the derivative does not exists on the interval given)

Derivative of this function

f'(x) = (1)x^-2

       = -2x^(-2-1)

       = -2x^(-3)

       = -2/x³

      =  -2/x³

f'(0) = -2/0 is undefined

Thus, mean value theorem can not be applied.

7 0
3 years ago
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