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3 years ago

When f(x)=4, what is th0

Mathematics

1 answer:

tiny-mole [99]3 years ago

3 0

Answer:

F(x)=4 means you plug in the 4 into where the variable is.

Step-by-step explanation:

Not really sure what your question meant but If you trying to find a problem with f(x)=4, then you plug in 4 into the variable of x.

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Solve the system of equation below by graphing them y= 2x-3 y=-2x+5

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Step-by-step explanation:

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3 years ago

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3 years ago

Which expression is equivalent to 3(5x-4)

erastovalidia [21]

Answer:

15x - 12

Step-by-step explanation:

3(5x - 4)

15x - 12 using the distributive property

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3 years ago

Read 2 more answers

Please help and thank you

ivann1987 [24]

First, let's try to isolate x to make the question easier to solve.

|2x-6|≤6

2x - 6 ≤ 6

2x - 6 ≥ -6

2x ≤ 12

x ≤ 6

2x ≥ 0

x ≥ 0

Knowing this, the graph should be equal or less than 6 and equal or greater than 0, making Option D the correct answer.

7 0

3 years ago

Find (a) the arc length and (b) the area of a sector.

Brut [27]

Answer:

a) 23.56 ft (2 dp)

b) 58.90 ft² (2 dp)

Step-by-step explanation:

Formula

$\textsf{Arc length}=r \theta$

$\textsf{Area of a sector}=\dfrac{1}{2}r^2 \theta$

$\quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle in radians)}$

Calculation

Given:

$\theta=\dfrac{3 \pi}{2}$
r = 5 ft

$\begin{aligned}\implies \textsf{Arc length} & =r \theta\\& = 5\left(\dfrac{3 \pi}{2}\right)\\& = \dfrac{15}{2} \pi \\& = 23.56\: \sf ft\:(2\:dp)\end{aligned}$

$\begin{aligned} \implies \textsf{Area of a sector}& =\dfrac{1}{2}r^2 \theta\\\\ & = \dfrac{1}{2}(5^2) \left(\dfrac{3 \pi}{2}\right)\\\\& = \dfrac{25}{2}\left(\dfrac{3 \pi}{2}\right)\\\\ & = \dfrac{75}{4} \pi \\\\& = 58.90 \: \sf ft^2\:(2\:dp)\end{aligned}$

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2 years ago

When f(x)=4, what is th0​

When f(x)=4, what is th0