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3 years ago

Find the length of segments AB and BC.

Mathematics

2 answers:

Kitty [74]3 years ago

6 0

Answer:

AB= 10

BC=48

Step-by-step explanation:

AB + BC= AC

(x-5) + (3x+3)= 6x-32

add like terms

x + 3x= 4x

-5+3= -2

now you have

4x-2= 6x-32

move like terms

subtract 4x on both sides

-2= 2x-32

add 32 to both sides

30= 2x

divide by 2

x=15

Plug 15 in for x and solve

ikadub [295]3 years ago

3 0

Answer:

AB = 10

BC = 48

Step-by-step explanation:

Assuming ABC is a straight line,

then AC = AB + BC

It is given that :

AB = x - 5

BC = 3x + 3

AC = 6x - 32

hence

AC = AB + BC

(6x - 32) = (x - 5) + (3x + 3)

6x - 32 = x - 5 + 3x + 3

6x - 32 = 4x - 2

6x - 4x = -2 + 32

2x = 30

x = 15

hence

AB = x - 5 = 15 - 5 = 10

BC = 3x + 3 = 3(15) + 3 = 48

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yaroslaw [1]

X=5
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here’s how i got it:
-6x+6y=6
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you will get this:

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Triangles ABC and DEF are similar triangles. What are the lengths of the unknown sides? Question 10 options: A) DE = 24 yd.; EF

Nataly_w [17]

Answer:

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Step-by-step explanation:

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tiny-mole [99]

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3 years ago

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fenix001 [56]

Answer:

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Step-by-step explanation:

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3 years ago

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Nastasia [14]

Answer:

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Step-by-step explanation:

$\frac{-\frac{1}{2}}{\frac{2\left(9+3\right)-4-3}{4\left(8\right)}}\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=\frac{-\frac{1}{2}}{\frac{2\left(9+3\right)-4-3}{4\cdot \:8}}\\\frac{2\left(9+3\right)-4-3}{4\cdot \:8}=\frac{17}{32}\\\frac{2\left(9+3\right)-4-3}{4\cdot \:8}\\2\left(9+3\right)-4-3=17\\2\left(9+3\right)-4-3\\2\left(9+3\right)=24\\2\left(9+3\right)\\\mathrm{Add\:the\:numbers:}\:9+3=12\\=2\cdot \:12\\\mathrm{Multiply\:the\:numbers:}\:2\cdot \:12=24\\=24-4-3$

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$\mathrm{Cancel\:the\:common\:factor:}\:2\\=-\frac{16}{17}\\\mathrm{Decimal:\quad }\:-0.94117$

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