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enot [183]
3 years ago
8

Which is true about the domain and range of each function? Both the domain and range change. Both the range and domain stay the

same. The domain stays the same, but the range changes. The range stays the same, but the domain changes.
Mathematics
2 answers:
Kryger [21]3 years ago
8 0

Answer:

B

Step-by-step explanation:

Just took it on edge

user100 [1]3 years ago
3 0

Answer:

D

The range stays the same, but the domain changes

Step-by-step explanation:

Just took the test

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2 ordered pairs of 6x+9y=0
jeka94

Answer:

1st

6x+9y=0

9y=6x

Y=6x/9

2nd

6x+9y =0

6x=9y

X=9y/6 ans

4 0
3 years ago
How many different ways are there to choose a subset of the set {1,2,3,4,5,6} so that the product of the members of the subset i
zvonat [6]

You have to pick at least one even factor from the set to make an even product.

There are 3 even numbers to choose from, and we can pick up to 3 additional odd numbers.

For example, if we pick out 1 even number and 2 odd numbers, this can be done in

\dbinom 31 \dbinom 32 = 3\cdot3 = 9

ways. If we pick out 3 even numbers and 0 odd numbers, this can be done in

\dbinom 33 \dbinom 30 = 1\cdot1 = 1

way.

The total count is then the sum of all possible selections with at least 1 even number and between 0 and 3 odd numbers.

\displaystyle \sum_{e=1}^3 \binom 3e \sum_{o=0}^3 \binom 3o = 2^3 \sum_{e=1}^3 \binom3e = 8 \left(\sum_{e=0}^3 \binom3e - \binom30\right) = 8(2^3 - 1) = \boxed{56}

where we use the binomial identity

\displaystyle \sum_{k=0}^n \binom nk = \sum_{k=0}^n \binom nk 1^{n-k} 1^k = (1+1)^n = 2^n

3 0
1 year ago
What decimal or fraction goes in the box
garik1379 [7]

Answer:

Your answer would be 7.

Step-by-step explanation:

Hope I helped!!!

7 0
3 years ago
This shape is made using three semicircles The smaller semicircle have diameters of 10cm Calculate the shaded area Take 3.142 an
CaHeK987 [17]

Answer:

The shaded area is 314.2 cm²

Step-by-step explanation:

Here we have the diameter, d₁ of the smaller semicircles as 10 cm

We note that the larger semicircle is subtended (bounded) by the two smaller semicircles;

1 shaded small semicircle and the other is blank

Therefore, the diameter, d₂ of the large semicircle = 10 + 10 = 20 cm

Also the area of the shaded figure consists of the removal of one small semicircle and the addition of the other semicircle to the area of the larger semicircle such that the area of the shaded figure is as follows

Shaded area of figure = π·d₂²/4 + π·d₁²/4 - π·d₁²/4 = π·d₂²/4 = π×20²/4 = 100×3.142 = 314.2 cm²

Shaded area = 314.2 cm².

6 0
3 years ago
Is △RWS ~ △QWT?<br><br> yes<br> no
vlabodo [156]
Yes, it is. they are.


8 0
3 years ago
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