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LenaWriter [7]
4 years ago
11

Let f be a function such that f(x)=2x-4 is defined on the domain.2 is less than or equal to x and x is lss than or equal to 6. T

he range of the function is
A) 0 is less thanor equal to and y is less than or equal to 8
B)0 is less than or equal to y and is less than infinity
C)2 is less than or equal to y andy is less than or equal to 6
D)negative infinity is less than y and y is less than infinity
Mathematics
2 answers:
Illusion [34]4 years ago
6 0

Answer:

Option A.

Step-by-step explanation:

The given function is  

f(x)=2x-4

It is defined on the domain 2\leq x\leq 6.

We need to find the range.

Range is the set of output values.

2\leq x\leq 6

Multiply 2 on each side.

4\leq 2x\leq 12

Subtract 4 from each side.

4-4\leq 2x-4\leq 12-4

0\leq f(x)\leq 8

0\leq y\leq 8

The range of the function is 0\leq y\leq 8.

Therefore, the correct option is A.

yulyashka [42]4 years ago
3 0
X  is <0 ; 6>
Y is ?
A > 0 and function is continuous and linear

Ymin = 2*2 - 4 = 0
Ymax = 2*6 - 4 = 8

Hence the answer is A) <0 ; 8>
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Answer:

120

Step-by-step explanation:

Since we're dealing with a problem where the order matters and the first two letters are already chosen we need to subtract the number of letters and the number of available slots per group.

We use the permutation formula to find the answer, but before that let's check values.

n = 8

k = 5

Now since there are two letters already chosen we have to deduct two from both the value of n and k.

n = 6

k = 3

Now we can use the permutation formula:

_{n}P_{k}=\dfrac{n!}{(n-k)!}

_{6}P_{3}=\dfrac{6!}{6-3)!}

_{6}P_{3}=\dfrac{6!}{3!}

_{6}P_{3}=\dfrac{6*5*4*3*2*1}{3*2*1}

The 3*2*1 cancels out and leaves us with:

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4 years ago
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Mekhanik [1.2K]

Answer:

98 degrees

Step-by-step explanation:

angle kj and lm r equal

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Use the Law of Cosines to find angle round to the nearest tenth if necessary).
ad-work [718]

Answer:

Step-by-step explanation:

3a)

find side HP  , let's set side HP = c in the law of cosines formula

then

c = sq rt [ 23^2 + 12^2 - 2*12*23*cos(117) ]

c = sq rt [ 529 + 144 - (-250.6027559) ]

c = sq rt [ 673 + 250.6027559 ]

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c = sq rt [ 9^2 + 22^2 - 2*9*22*cos(59) ]

c = sq rt [ 81 + 484 -396*0.5150380 ]

c = sq rt [ 565 - 203.955 ]

c = sq rt [ 391.04495]

c = 19.7748

PR = 19.8  ( rounded to nearest 10th )

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they want us to find ∠B,   in the formula for cosines ∠B is really ∠C, it's a bit confusing.. I'm sure they did that on purpose to mess people up.  Kind of underhanded  :/    Tell the teacher that you're "appalled" at the deviousness of the test  :/   stage a protest against math that teaches treachery  :P  storm the principals/ dean's  office with paper airplanes and pictures of anime   :DDD  okay back to math.

a=28

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B = arcCos [28^2 + 15^2 - 17^2 / 2*28*15]

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3d)

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∠A = arcCos[604/672]

∠A = arcCos[0.8988095]

∠A =25.99797 °

∠A =26.0 °  ( rounded to nearest 10th)

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