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S_A_V [24]
3 years ago
5

Find the range of the function y=1/2x+3 when the domain is (-2, 0, 2, 4)

Mathematics
2 answers:
Aleks04 [339]3 years ago
7 0

Answer:

domain=2

Step-by-step explanation:

-2+2=0 0+2=2 2+2=4

goblinko [34]3 years ago
7 0

Answer: The range of values for the function is (2,4,3.3,3.1)

Step-by-step explanation:

Given the expression y=1/2x+3 in the domain (-2, 0, 2, 4)

Step one : for x= - 2 we have

y=1/(2*-2)+3

y=1/-4+3

y=1/-1+3

y=-1+3

y=2

Step two :

If x is 0 we have

y=1/2*0+3

y=1/0+3

y=1+3

y=4

step three :

if x is 2

y=1/2*2+3

y=1/4+3

y=13/4

y=3.25

y =3.3

Step four :

if x is 4

y=1/2*4+3

y=1/8+3

y=25/8

y=3.1

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How many 3/4 cups of milk would be needed to make 21 cups of milk for a large recipe?

4 0
3 years ago
50 POINTS!!!!!
Makovka662 [10]

We can adjust the data by adding 4 to everything before we calculate the statistics.  Or we can calculate the statistics on the given data and just add 4 to everything at the end.  We'll get the same answer either way.

Let's sort the seven data points:  5 5 5 7 7 9 10

Those add up to 48 so the mean is 48/7 = 6.9

The one in the middle is 7 so the median = 7

The mode is the most common one, mode = 5

The range is the difference between max and min, so range = 10 - 5 = 5

In the second week we add four to everything.  Since that adds four to the min and max, the range doesn't change.

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3 years ago
Can someone explain to me why you flip the reciprial
puteri [66]

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Step-by-step explanation:

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3 years ago
Find the equation of the line that is parallel toy=32x−5
Alona [7]

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Step-by-step explanation:

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8 0
3 years ago
In rectangle abcd, points p and q lie on side AB and DC respectively. Angle PMQ is a right angle, M is the midpoint of side BC a
nirvana33 [79]

Answer:

PM:MQ = 8:3.

Step-by-step explanation:

\rm \angle B\hat{M}P + 90^{\circ} + \angle C\hat{M}Q = 180^{\circ};

\implies \rm \angle B\hat{M}P + \angle C\hat{M}Q = 90^{\circ};

\implies \rm 90^{\circ} - \angle B\hat{M}P = \angle C\hat{M}Q.

Also,

\rm \angle B\hat{P}M = 90^{\circ} - \angle B\hat{M}P in right triangle PBM.

Thus \rm \angle{P\hat{B}M} = \angle C\hat{M}Q.

Additionally \rm \angle \hat{B} = 90^{\circ} = \angle \hat{C}.

Therefore \rm \triangle PBM \sim \triangle MCQ.

\rm \displaystyle BC = 2\;MC for M is the midpoint of segment BC.

\rm \displaystyle PB = \frac{4}{3}BC = \frac{8}{3}MC.

\rm \triangle PBM \sim \triangle MCQ implies that

\displaystyle \rm PM:MQ = PB:MC = 1:\frac{8}{3} = 8:3.

8 0
3 years ago
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