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fenix001 [56]
4 years ago
11

A circle has a radius of 1 cm.

Mathematics
1 answer:
Bezzdna [24]4 years ago
3 0
A, the circumference of a circle is greater than the area.
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A computer technician keeps track of his earning through tout each months. The technician observes that his earnings are a linea
notka56 [123]

Answer:

Part A :  E = 35h + 200

Part B : The per hour charges of the technician is $35/hour.

Step-by-step explanation:

The technician observes that his earnings are a linear function of the number of hours he works during the month. The technician finds that when he works 55 hours during the month, he earns $2,125 and when he works 30 hours, he earns $1,250 .

If we consider the number of hours that he works in a month is h and the amount he earns in dollars is E, then (30,1250) and (55,2125) are the two ordered pairs.

Part A : Therefore, the linear function to model the relationship between the number of hours worked and the money earned will be

\frac{E - 2125}{2125 - 1250} = \frac{h - 55}{55 - 30}

⇒ E - 2125 = 35(h - 55)

⇒ E - 2125 = 35h - 1925

⇒ E = 35h + 200

Part B : The equation above is in the slope-intercept form and the slope is 35 which means that the per hour charge of the technician is $35/hour. (Answer)

4 0
3 years ago
Which option best defines parallel lines?
Harlamova29_29 [7]
Parallel lines are coplanar lines (lines that lie on the same plane) and they do not intersect.

It’s important to understand that they remain the same distance apart over their entire lengthy which means they never get closer together of farther away from each other.
7 0
4 years ago
26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+
Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $ \forall (a, b) \in \mathbb{R}^2 $, $ (a, b) R(a, b) $.

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$ (a, b) \sim (a, b) $ since $ a^2 + b^2 = a^2 + b^2 $

This is true for any pair of numbers in $ \mathbb{R}^2 $. So, $ \sim $ is reflexive.

Symmetry:

$ \sim $ is symmetry iff whenever $ (a, b) \sim (c, d) $ then $  (c, d) \sim (a, b) $.

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$ a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13 $

$ c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42 $

Hence, $ (a, b) \sim (c, d) $ since $ a^2 + b^2 \leq c^2 + d^2 $

Note that $ c^2 + d^2 \nleq a^2 + b^2 $

Hence, the given relation is not symmetric.

Transitive:

$ \sim $ is transitive iff whenever $ (a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f) $ then $ (a, b) \sim (e, f) $

To prove transitivity let us assume $ (a, b) \sim (c, d) $ and $ (c, d) \sim (e, f) $.

We have to show $ (a, b) \sim (e, f) $

Since $ (a, b) \sim (c, d) $ we have: $ a^2 + b^2 \leq c^2 + d^2 $

Since $ (c, d) \sim (e, f) $ we have: $ c^2 + d^2 \leq e^2 + f^2 $

Combining both the inequalities we get:

$ a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2 $

Therefore, we get:  $ a^2 + b^2 \leq e^2 + f^2 $

Therefore, $ \sim $ is transitive.

Hence, proved.

3 0
3 years ago
____% of 50 = 30<br> Please help
aleksandr82 [10.1K]

Answer:

60%

Step-by-step explanation:

30 is 60 percent of 50

5 0
3 years ago
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Given m ||n, find the value of x. (x+14)° (4x-9)°​
deff fn [24]

Answer:4.5 hope this helps

Step-by-step explanation:

6 0
3 years ago
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