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gtnhenbr [62]
1 year ago
9

Identify the function that contains the data in the following table:Х-20o235f(x)531224O f(x) = x/ + 1O f(x) = x - 21O f(x) = lx

- 21 - 1Of(x) = lx - 21+1

Mathematics
1 answer:
kirill [66]1 year ago
3 0

To identify the function that contains the data in the table, we should first visualize the data.

A graph of the data is shown below:

From the plot above, we can identify that:

The graph above is a graph of f(x) = |x|, translated to the right 2 units and translated upwards 1 unit.

Hence, the function is:

f(x)\text{  = |x-2| + 1}

Answer:

Option D

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42=Wf×64 (what fraction)
Dmitry [639]
So basically 42=x times 64 or 42=64x where you want x to be represented as a fraction so just don't simplify as much

42=64x
divide both sdies by 64
42/64=64/64x
42/64=1x
x=42/64
simplify
42/64=21/32

the answer is 21/32 in simplest form
7 0
3 years ago
20 POINTS!!!
notka56 [123]

\dfrac{2}{\sqrt3\cos x+\sin x}=\sec\left(\dfrac{\pi}{6}-x\right)\\\\\dfrac{2}{\sqrt3\cos x+\sin x}=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}\ \ \ \ \ (*)\\----------------\\\\\cos\left(\dfrac{\pi}{6}-x\right)\ \ \ \ |\text{use}\ \cos(x-y)=\cos x\cos y+\sin x\sin y\\\\=\cos\dfrac{\pi}{6}\cos x+\sin\dfrac{\pi}{6}\sin x=\dfrac{\sqrt3}{2}\cos x+\dfrac{1}{2}\sin x=\dfrac{\sqrt3\cos x+\sin x}{2}\\------------------------------

(*)\\R_s=\sec\left(\dfrac{\pi}{6}-x\right)=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}=\dfrac{1}{\dfrac{\sqrt3\cos x+\sin x}{2}}\\\\=\dfrac{2}{\sqrt3\cos x+\sin x}=L_s

6 0
3 years ago
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What’s the answer <br><br> 1/6+1/2
miskamm [114]

Answer:

4/6 or 2/3

Step-by-step explanation:

Using Common denominators:

3/6 = 1/2

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This can reduce to 2/3

8 0
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Please help! I’ll give brainliest
Vilka [71]

Answer:

Step 1: Simplify both sides of the equation.

−7f=12

Step 2: Divide both sides by -7.

−7f

−7

=

12

−7

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Step-by-step explanation

sorry for the answer like this i couldnt find the fractions but the answer is f=-12/7 sorry :(

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Where the image has the opposite orientation as the preimage
Elena L [17]
Im not in this math section yet but i think that you can look it up as a worksheet and find the answer on google.
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