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2 years ago

100 points but please don’t post a very useless answers! >:(

Mathematics

1 answer:

Lelu [443]2 years ago

7 0

Answer:

See below.

Step-by-step explanation:

54.

The radius of the small circles is x.

The radius of the large circle is 2x.

Area of large circle:

πr² = π(2x)² = π4x² = 4πx²

Area of 1 small circle:

πr² = πx²

Area of 2 small circles:

2πx²

shaded area = area of large circle - area of 2 small circles

shaded area = 4πx² - 2πx²

shaded area = 2πx²

57.

The radius if the circle is x√2.

The side of the square is 2x.

Area of circle:

πr² = π(x√2)² = 2πx²

Area of square:

s² = (2x)² = 4x²

shaded area = area of circle - area of square

shaded area = 2πx² - 4x² or (2π - 4)x²

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Sarita showed the steps to solving the equation.

shtirl [24]

The option that explains Sarita's mistake is Sarita’s solution is correct. She made an error in her verification.

<h3>What was Sarita's mistake?</h3>

The given equation is: 5(x - 3) + 7(x + 4) = 73

In order to determine the value of x, take the following steps:

1. Apply the Distributive property:

5x - 15 + 7x + 28 = 73

2. Add similar terms together:

12x + 13 = 73

3. Combine similar terms:

12x = 73 - 13

4. Add similar terms

12x = 60

5. Divide both sides of the equation by 12

x = 60 / 12

x = 5

In order to verify the answer, substitute for x in the given equation :

5(5 - 3) + 7(5 + 4) = 73

5(2) + 7(9) = 73

10 + 63 = 73

73 = 73

Please find attached the complete question. To learn more about equations, please check: brainly.com/question/14446120

#SPJ1

6 0

1 year ago

Can anyone please help me with this!!! :(

UkoKoshka [18]

Answer:

ano ba gagawin

parang Ang Dali pero ano gagawin dyan

3 0

3 years ago

Read 2 more answers

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,