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2 years ago

I need to find A and C

Mathematics

2 answers:

Novay_Z [31]2 years ago

7 0

if we take a look of the graph, let's notice that

AC = 4x + 4

AB = 14

BC = 3x - 4

now, AC = AB + BC,

$\bf AC=AB+BC\implies AC-BC=AB\implies \stackrel{AC}{(4x+4)}-\stackrel{BC}{(3x-4)}=\stackrel{AB}{14}
\\\\\\
4x+4-3x+4=14\implies x+8=14\implies x=6
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
AC=4(6)+4\implies \boxed{AC=28}$

k0ka [10]2 years ago

6 0

Then find it sksngjsjjsjdjgjsjajcjvjsjajf

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Charra [1.4K]

Answer:

$2a^3b^2\sqrt[3]{3a}$

Step-by-step explanation:

Use the following rules for exponents:

$a^m*a^n=a^{m+n}\\\\\sqrt[3]{x^3}=x$

Simplify 24. Find two factors of 24, one of which should be a perfect cube:

$8*3=24\\\\2^3=8$

Insert:

$\sqrt[3]{2^3*3a^{10}b^6}$

Now split the exponents. Split 10 into as many 3's as possible:

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Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^6}$

Split 6 into as many 3's as possible:

$6=3+3$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^3*b^3}$

Now simplify. Any terms with an exponent of 3 will be moved out of the radical (rule #2):

$2\sqrt[3]{3*a^3*a^3*a^3*a^1*b^3*b^3}\\\\\\2*a*a*a\sqrt[3]{3*a^1*b^3*b^3}\\\\\\2*a*a*a*b*b\sqrt[3]{3*a^1}$

Simplify:

$2a^3b^2\sqrt[3]{3a}$

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