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Tanya [424]
3 years ago
7

Can someone explain how to solve

Mathematics
1 answer:
wel3 years ago
4 0

The x-intercept is the number without the variable.

So, #1's correct answer would be A. because -10 is not greater than -2.


You might be interested in
(-10) - |(-20) – (+30) |
myrzilka [38]

ANSWER: -60

EXPLANATION: To solve for (-10) - |(-20) – (+30) |, we need to do:

−10−| −20 − 30|, which equals −10−|−50|. Next, we do −10−50, which is -60.

−10−|−20−30|

= −10−|−50|

= −10−50

= −60

6 0
2 years ago
Tehara has read 115 pages of a 305 page book she reads 10 pages each day how many days will it take to finish
scZoUnD [109]

Answer: 19 days

Step-by-step explanation:

305-115=190

190 divided by 10= 19

7 0
3 years ago
Read 2 more answers
What is 0.95 rounded off to the nearest whole number
mihalych1998 [28]

1.00 would be your answer because 95 rounded up is 100. So, 0.95 is closest to 1.00

8 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
How to expand (2x+1)(x+1)
Inga [223]
(2x+1)(x+1)
Now multiply 2x to both x and +1
2x^2+2x
Do the same with 1 multiply it to x and +1
2x^2+2x +x+1
Simplify
2x^2+3x+1
8 0
3 years ago
Read 2 more answers
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