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Lunna [17]
2 years ago
13

I will also give brainliest on this one What’s the surface area of this round to the tenth if you have to

Mathematics
1 answer:
Natali5045456 [20]2 years ago
5 0
542.4
I believe is the answer
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Exam pls help 34 points
FrozenT [24]

Answer:

it would be 4/20 I thinkk

Step-by-step explanation:

4 0
3 years ago
given examples of relations that have the following properties 1) relexive in some set A and symmetric but not transitive 2) equ
rodikova [14]

Answer: 1) R = {(a, a), (а,b), (b, a), (b, b), (с, с), (b, с), (с, b)}.

It is clearly not transitive since (a, b) ∈ R and (b, c) ∈ R whilst (a, c) ¢ R. On the other hand, it is reflexive since (x, x) ∈ R for all cases of x: x = a, x = b, and x = c. Likewise, it is symmetric since (а, b) ∈ R and (b, а) ∈ R and (b, с) ∈ R and (c, b) ∈ R.

2) Let S=Z and define R = {(x,y) |x and y have the same parity}

i.e., x and y are either both even or both odd.

The parity relation is an equivalence relation.

a. For any x ∈ Z, x has the same parity as itself, so (x,x) ∈ R.

b. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R.

c. If (x.y) ∈ R, and (y,z) ∈ R, then x and z have the same parity as y, so they have the same parity as each other (if y is odd, both x and z are odd; if y is even, both x and z are even), thus (x,z)∈ R.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial but not transitive, so the relation provided in (1) satisfies this condition.

Step-by-step explanation:

1) By definition,

a) R, a relation in a set X, is reflexive if and only if ∀x∈X, xRx ---> xRx.

That is, x works at the same place of x.

b) R is symmetric if and only if ∀x,y ∈ X, xRy ---> yRx

That is if x works at the same place y, then y works at the same place for x.

c) R is transitive if and only if ∀x,y,z ∈ X, xRy∧yRz ---> xRz

That is, if x works at the same place for y and y works at the same place for z, then x works at the same place for z.

2) An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial and not transitive.

QED!

6 0
3 years ago
On a coordinate plane, a line is drawn from Rock to Tree. The x- and y-axes are labeled feet. Rock is at (3, 2) and Tree is at (
-Dominant- [34]

Answer:

The second choice (7.6,8.8)

Step-by-step explanation:

The missing image is attached

We use the section formula:

(x,y)=(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})

We substitute x_1=3,x_2=16,y_1=2,y_2=21 and m=5, n=9

(x,y)=(\frac{5*16+9*3}{5+9},\frac{5*21+5*2}{9+5})

We simplify to get:

(x,y)=(\frac{80+27}{14},\frac{105+10}{14})

This simplifies to:

(x,y)=(\frac{107}{14},\frac{125}{14})

In decimals we get:

(x,y)=(7.6,8.8)

6 0
2 years ago
Read 2 more answers
Find the exact surface area of the figure.
puteri [66]
The surface area is A = 297.33
7 0
2 years ago
The dimensions of a rectangular garden were 5m by 12m. Each dimensions was increased by the same amount. the garden has an area
Alex_Xolod [135]
Let's say, the amount that each side was increased was "a"
so, it was 5 wide and 12 long, now it's 5+a wide and 12+a long

we know the area is 120, so whatever "a" is, we know that

(5+a)(12+a) = 120

so   \bf (5+a)(12+a) = 120\implies 60+17a+a^2=120
\\\\\\
a^2+\underline{17} a\underline{-60}=0\qquad \qquad 
\begin{cases}
20\cdot -3\implies \underline{-60}\\\\
20-3\implies \underline{17}
\end{cases}

use those factors, solve for "a"
4 0
3 years ago
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