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Sergio039 [100]
2 years ago
15

Antwan determines the distance between the points –7 and 2 on a number line. Maggie determines the difference between the number

s –7 and 2. How are Antwan’s and Maggie’s solutions related?
Maggie’s solution is the absolute value of Antwan’s solution.

Antwan’s solution is the absolute value of Maggie’s solution.

Both solutions are greater than either of the two numbers in the problem.

Both solutions are less than either of the two numbers in the problem.
Mathematics
1 answer:
-Dominant- [34]2 years ago
6 0

Antawan's solution is the absolute value of Maggie's solution is the solution related to Antawan's and Maggie's solution.

Given

Antawan determines the distance between two points as 9.

Maggie finds the difference between two numbers ₋ 7 ₋ 2 = ₋9

Hence we notice that Antawan's solution is the absolute value to the Maggie's solution.

The non-negative value of x, regardless of its sign, is the absolute value (or modulus) | x | of a real number x. For instance, 5 has an absolute value of 5 and so does 5, which likewise has an absolute value of 5. One way to conceptualize a number's absolute value is as its separation from zero on the real number line.

hence option 2 is right.

Learn more about absolute values here:

brainly.com/question/5012769

#SPJ9

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GIVING OUT BRAINLIEST TO WHOEVER GETS ALL OF THEM RIGHT
Thepotemich [5.8K]

Answer:

4) \frac{x}{7\cdot x +x^{2}} is equivalent to \frac{1}{7+x} for all x \ne -7. (Answer: A)

5) \frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}} is equivalent to -\frac{14}{1-5\cdot x} for all x \ne \frac{1}{5}. (Answer: B)

6) \frac{x+7}{x^{2}+4\cdot x - 21} is equivalent to \frac{1}{x-3} for all x \ne 3. (Answer: None)

7) \frac{x^{2}+3\cdot x -4}{x+4} is equivalent to x - 1. (Answer: None)

8)  \frac{2}{3\cdot a}\cdot \frac{2}{a^{2}} is equivalent to \frac{4}{3\cdot a^{3}} for all a\ne 0. (Answer: A)

Step-by-step explanation:

We proceed to simplify each expression below:

4) \frac{x}{7\cdot x +x^{2}}

(i) \frac{x}{7\cdot x +x^{2}} Given

(ii) \frac{x}{x\cdot (7+x)} Distributive property

(iii) \frac{1}{7+x} \cdot \frac{x}{x} Distributive property

(iv) \frac{1}{7+x} Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

7+x = 0

x = -7

Hence, we conclude that \frac{x}{7\cdot x +x^{2}} is equivalent to \frac{1}{7+x} for all x \ne -7. (Answer: A)

5) \frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}

(i) \frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}} Given

(ii) \frac{x^{3}\cdot (-14)}{x^{3}\cdot (1-5\cdot x)} Distributive property

(iii) \frac{x^{3}}{x^{3}} \cdot \left(-\frac{14}{1-5\cdot x} \right) Distributive property

(iv) -\frac{14}{1-5\cdot x} Commutative property/Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

1-5\cdot x = 0

5\cdot x = 1

x = \frac{1}{5}

Hence, we conclude that \frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}} is equivalent to -\frac{14}{1-5\cdot x} for all x \ne \frac{1}{5}. (Answer: B)

6) \frac{x+7}{x^{2}+4\cdot x - 21}

(i) \frac{x+7}{x^{2}+4\cdot x - 21} Given

(ii) \frac{x+7}{(x+7)\cdot (x-3)} x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})

(iii) \frac{1}{x-3}\cdot \frac{x+7}{x+7} Commutative and distributive properties.

(iv) \frac{1}{x-3} Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

x-3 = 0

x = 3

Hence, we conclude that \frac{x+7}{x^{2}+4\cdot x - 21} is equivalent to \frac{1}{x-3} for all x \ne 3. (Answer: None)

7) \frac{x^{2}+3\cdot x -4}{x+4}

(i) \frac{x^{2}+3\cdot x -4}{x+4} Given

(ii) \frac{(x+4)\cdot (x-1)}{x+4}  x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})

(iii) (x-1)\cdot \left(\frac{x+4}{x+4} \right) Commutative and distributive properties.

(iv) x - 1 Existence of additive inverse/Modulative property/Result

Polynomic function are defined for all value of x.

\frac{x^{2}+3\cdot x -4}{x+4} is equivalent to x - 1. (Answer: None)

8) \frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}

(i) \frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}

(ii) \frac{4}{3\cdot a^{3}} \frac{a}{b}\cdot \frac{c}{d} = \frac{a\cdot b}{c\cdot d}/Result

Rational functions are undefined when denominator equals 0. That is:

3\cdot a^{3} = 0

a = 0

Hence, \frac{2}{3\cdot a}\cdot \frac{2}{a^{2}} is equivalent to \frac{4}{3\cdot a^{3}} for all a\ne 0. (Answer: A)

6 0
3 years ago
Please Help Me ASAP Will Thank Rate 5 Starts And Mark Brainliest If You Are CORRECT
Ivan

Answer:

The correct option is 4.

Step-by-step explanation:

It is given that figure ABCD is transformed to figure A′B′C′D′.

The vertices of ABCD are A(1,1), B(3,3), C(4,2) and D(4,1).

The vertices of A'B'C'D' are A'(1,-3), B'(3,-1), C'(4,-2) and D'(4,-3).

It is clear that the figure translate 4 units down and the rule of translation is

Translation is a rigid transformation, it means image and preimge are congruent.

Since ABCD is translate 4 units down to make A′B′C′D′, therefore

Congruent part of congruent figures are congruent.

                 

Therefore the correct option is 4.

I Hope It's Helpful :)

4 0
3 years ago
Read 2 more answers
Pls help me I would dearly appreciate it!
TiliK225 [7]

Answer:

1. c

2. c

Step-by-step explanation:

1. subract 8 from both sides to get b/-4=-3

multiply both sides by -4 to isolate b

you get b=12

2. combine like terms 3c-4=-8+2c

add 4 to both sides: 3c = -4+2c

subtract 2c from both sides

c=-4

8 0
3 years ago
Read 2 more answers
If the area to the right of x in a normal distribution is 0.65, what is the area to the left of x? *
Nutka1998 [239]
The first option 0.65 might be right
5 0
3 years ago
They are separate questions.
NNADVOKAT [17]

Answer: S+2,  

a+b,  

c-(a+b),

z+12,

t- (s+5) ,

(15+d) -c

Step-by-step explanation:

4 0
3 years ago
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