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Mashutka [201]
2 years ago
13

Somebody pls help me with this question pls

Mathematics
1 answer:
monitta2 years ago
4 0
292in^3
3x6x10=180 because I’ve split the shape in half
7x4x4=112
180+112=292
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spin [16.1K]

Answer:

the answer is b

Step-by-step explanation:

first you simplify the numbers by 25 and end up with x - 5 over x^2 and then you simplify by x and you get -5 over x

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3 years ago
I need help, the solid below is made from cubes find it volume
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Answer:

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2 years ago
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Consider functions f and g.1 + 12f(1) = 12 + 4. – 12for * # 2 and 7 -64.2 – 16. + 1641 +48for a # -12 Which expression is equal
lisabon 2012 [21]

Given the following functions below,

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}\text{ and} \\ g(x)=\frac{4x^2-16x+16}{4x+48} \end{gathered}

Factorising the denominators of both functions,

Factorising the denominator of f(x),

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}=\frac{x+12}{x^2+6x-2x-12}=\frac{x+12}{x(x+6)-2(x+6)}=\frac{x+12}{(x-2)(x+6)} \\ f(x)=\frac{x+12}{(x-2)(x+6)} \end{gathered}

Factorising the denominator of g(x),

\begin{gathered} g(x)=\frac{4x^2-16x+16}{4x+48}=\frac{4(x^2-4x+4)}{4(x+12)} \\ \text{Cancel out 4 from both numerator and denominator} \\ g(x)=\frac{x^2-4x+4}{x+12}=\frac{x^2-2x-2x+4}{x+12}=\frac{x(x-2)-2(x-2)}{x+12}=\frac{(x-2)^2}{x+12} \\ g(x)=\frac{(x-2)^2}{x+12} \end{gathered}

Multiplying both functions,

undefined

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1 year ago
A number x is greater than 0. write the word sentence as an inequality
IgorLugansk [536]

9514 1404 393

Answer:

  x > 0

Step-by-step explanation:

The only thing that needs translation to symbols is "greater than". Your "word sentence" already uses the symbols x and 0. The symbol for greater than is >. Your inequality is ...

  x > 0

8 0
3 years ago
PLEASE I NEED HELP QUICKLY!!! Find the solution for the given system of equations in the form (x,y).
goblinko [34]

Answer:

  • (4, - 4)

Step-by-step explanation:

<u>Given system:</u>

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<u>Multiply the first equation by 7 and add up the equations:</u>

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  • 10y = 40
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<u>Find y:</u>

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3 0
2 years ago
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