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Artemon [7]
3 years ago
5

What is the range of the relation?

Mathematics
1 answer:
Rasek [7]3 years ago
8 0

Answer:

the range of the relation is 3

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What is 2+8? Help needed!!!!!!!!
denpristay [2]
10......................:)
4 0
3 years ago
What is the area? Please explain steps so I can attempt to understand
yarga [219]

Answer:

12 \sqrt{2}

Step-by-step explanation:

you can use hypotenuse formula

45-45-90 isosceles right triangle

so, 45= x

x² + x² = 24²

2x² = 24²

x²= 24.12

x= \sqrt{288}  = 12 \sqrt{2}

or :

you can use this rule: if you see a 45 45 90 isosceles right triagngle (you can look picture) 45=a cm and 90 = a\sqrt{2}

cm

so ;

a \sqrt{2}  = 24

a= 12 \sqrt{2}

hope this helps ^-^

3 0
4 years ago
Suppose S and T are mutually exclusive events. Find P(S or T). P(S) = 20%, P(T) = 22%
Svetradugi [14.3K]
S and P are mutually exclusive events.
P ( S ) = 20% = 0.20
P ( T ) = 22 % = 0.22
The probability of mutually exclusive events:
P ( S or T ) = P ( S ) + P ( T )
P ( S or T ) = 0.20 +0.22 = 0.42
Answer: The probability of ( S or T ) is 0.42 or 42 %. 
8 0
3 years ago
Read 2 more answers
In the figure shown, lines 1 and 2 are parallel. The figure shows two diagonal parallel lines Italic l 1 and Italic l 2 and a ho
horsena [70]
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5 0
3 years ago
Acircular ring has across -section as shown .If the outer radias 20 mm and the inner radius 16 mm .Calculate the cross - section
Zina [86]

Answer:

452.45mmm^2

Step-by-step explanation:

Step one:

Given data

outer radius R= 20mm

inner radius r= 16mm

let us find the area of the outer circle

Area= πR^2

Area= 3.142*20^2

Area= 3.142*400

Area= 1256.8 mm^2

let us find the area of the inner circle

Area= πr^2

Area= 3.142*16^2

Area= 3.142*256

Area= 804.35 mm^2

Hence the area of the section is

Area of outer- Area of inner

=1256.8-804.35

=452.45mmm^2

4 0
3 years ago
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