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2 years ago

Simplify the expression 2 × 36 − 24 ÷ 6

Mathematics

2 answers:

natka813 [3]2 years ago

7 0

Answer:

68

Step-by-step explanation:

*Remember the order of operations PEMDAS*(Parentheses, Exponents, Multiplication and Division, Addition and Subtraction)

2 x 36 = 72
24 / 6= 4
now, 72 - 4= 68

hope it helps:)

marin [14]2 years ago

3 0

Answer:

Step-by-step explanation:

P- parentheses

E- exponential

M- multiplication

D- division

A- addition

S- subtraction

2 x 36 = 72

72 - 24 ÷ 6

24 ÷ 6 = 4

72 - 4 = 68

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The function d(x) = |x 100| can be used to find the number of degrees a container of water at a temperature of x degrees Celsius

Volgvan

Answer:

$f(d(x)) = \frac{9}{5}|x^{100}| + 32$

Step-by-step explanation:

The function d(x) takes a value of x in degrees centigrade and provides the number of degrees that a container of water at that temperature x is far from the boiling point of water.

The function f(d) takes a value d in degrees centigrade and returns a value d(x) in degrees fahrenheit.

Therefore, by doing f(d(x)) we are introducing the function d(x) within the function f(d).

So the range of d(x) now is the domain of f(d(x))

This means that the function f(d(x)) shows the number of degrees Fahrenheit that a water container at a temperature x in degrees Celsius is far from the boiling point of water.

$f(d(x)) = \frac{9}{5}|x^{100}| + 32$

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3 years ago

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Which term best describes the statement given below?

GarryVolchara [31]

B is as it is a logical form of reasoning. if a=50, b=30, c=20
if a > b and b > c then a > c

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3 years ago

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Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.

Eduardwww [97]

Answer:

See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

\frac{x+1}{x-1} = \frac{y+1}{y-1}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

Hence f is one - one

let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}

\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )

\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}

\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

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Step-by-step explanation:

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