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blondinia [14]
3 years ago
13

If anyone is good at geometry, pls help :)

Mathematics
1 answer:
Vadim26 [7]3 years ago
4 0

Answer:

<u>Segment</u><u> </u><u>Addition</u><u> </u><u>Postulate</u>

<u>BC</u> + <u>AB</u> = <u>CA</u> [BOTH answers]

<u>Application</u><u> </u><u>with</u><u> </u><u>Midpoint</u>

<u>BC</u> = <u>BA</u>

1a) 9 = <em>x</em>

1b) 34 = FA

1c) 3 = AT

2a) 11 = <em>x</em>

2b) 37 = BO

2c) 37 = OX

2d) 74 = BX

Step-by-step explanation:

The Segment Addition Postulate is basically saying that if

___

point <em>B</em><em> </em>is in between <em>AC</em><em> </em>[NOT<em> </em>the <em>midpoint</em><em> </em>(or halfway)],<em> </em>then both segments [AB and BC] would make up the whole thing [CA, or vice versa]. They are just explaining it in two unique ways you can word it.

The Midpoint Application tells you that if point <em>B</em><em> </em>is a

___

<em>midpoint</em><em>,</em><em> </em>meaning <em>B</em> is halfway in between <em>AC</em><em>,</em><em> </em>then both segments [AB and BC] are congruent to each other.

1a) 37 = [2<em>x</em> - 15] + [3<em>x</em> + 7]

37 = 5<em>x</em> - 8

+ 8 + 8

____________

45 = 5<em>x</em>

___ ___

5 5

9 = <em>x</em>

1b) Solving for FA

3(9) + 7 = 27 + 7 = 34

1c) Solving for AT

2(9) - 15 = 18 - 15 = 3

This obviously had to be the answer because 3 + 34 = 37.

2a) 2<em>x</em> +15 = 4<em>x</em> - 7

- 4<em>x</em> - 4<em>x</em>

_________________

-2<em>x</em> + 15 = -7

- 15 -15

_____________

-2<em>x</em> = -22

<em>x</em><em> </em>= 11

2b) Solving for BO

4(11) - 7 = 44 - 7 = 37

2c) Solving for OX

2(11) + 15 = 22 + 15 = 37

2d) Solving for BX

2 \times 37 = 74

I am joyous to assist you anytime.

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there are 104 red candies in a candy machine. this represents 26% of the candy in the machine. how many total are in the candy m
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There are 400 candies in the machine.

Step-by-step explanation:

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x be the total candies in the machine.

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Answer image is attached.

Step-by-step explanation:

Given rational expressions:

1.\ \dfrac{x^2+x+4}{x-2}\\2.\ \dfrac{x^2-x+4}{x-2}\\3.\ \dfrac{x^2-4x+10}{x-2}\\4.\ \dfrac{x^2-5x+16}{x-2}

And the rewritten forms:

(x-2)+\dfrac{6}{x-2}\\(x+3)+\dfrac{10}{x-2}\\(x+1)+\dfrac{6}{x-2}\\(x-3)+\dfrac{10}{x-2}

We have to match the rewritten terms with the given expressions.

Let us consider the rewritten terms and let us solve them one by one by taking LCM.

(x-2)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x-2)^{2}+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+4+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+10}{x-2}

So, correct option is 3.

(x+3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x+3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{(x^2+3x-2x-6)+10}{x-2}\\\Rightarrow \dfrac{x^2+x+4}{x-2}

So, correct option is 1.

(x+1)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x+1)(x-2)+6}{x-2}\\\Rightarrow \dfrac{x^{2} +x-2x-2+6}{x-2}\\\Rightarrow \dfrac{x^{2} -x+4}{x-2}

So, correct option is 2.

(x-3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x-3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{x^2-3x-2x+6+10}{x-2}\\\Rightarrow \dfrac{x^2-5x+16}{x-2}

So, correct option is 4.

The answer is also attached in the answer area.

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2 years ago
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