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makkiz [27]
3 years ago
6

Use the given graph to determine the limit, if it exists.

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
6 0
As x approaches 3 from the right, the limit is 3.  There is a closed hole up there AT x=3 y=7, but we don't care about what happens AT x=3, only what happens as x APPROACHES 3. 
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For the eradication of malignancies like the aggressive type of human brain tumour known as glioblastoma multiforme (GBM), 3D platforms are crucial for tracking tumour growth and assessing treatment candidates.

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4 0
2 years ago
How would I find the diameter?
adell [148]
By measuring the width of the circle or sphere radius is half of the diameter
3 0
3 years ago
100 POINTS!! MICROSOFT EXCEL
murzikaleks [220]

Step-by-step explanation:

A new coffee shop is being built. Its location is the reflection of the arcade's coordinates across

the y-axis. Which procedure will find the correct distance between the arcade and the new coffee shop

4 0
2 years ago
Read 2 more answers
Kaylee found the surface area, in square inches of a rectangular prism by using formula. 2(5×3)+2(5×8)+2(3×8) What is the surfac
sladkih [1.3K]

Answer:

158

Step-by-step explanation:

Simplify the following:

2×5×3 + 2×5×8 + 2×3×8

5×3 = 15:

2×15 + 2×5×8 + 2×3×8

5×8 = 40:

2×15 + 2×40 + 2×3×8

3×8 = 24:

2×15 + 2×40 + 2×24

2×15 = 30:

30 + 2×40 + 2×24

2×40 = 80:

30 + 80 + 2×24

2×24 = 48:

30 + 80 + 48

| 8 | 0

| 4 | 8

+ | 3 | 0

1 | 5 | 8:

Answer:  158

6 0
3 years ago
Resolve to me these exercises please?
Dima020 [189]
A)4^(n+3)=8^14
2^(2×(n+3))=2^(3×14)
2^(2n+6)=2^42
2^2n=2^36
n=18

b) (assuming a : is divide)
3^(2n+1)=9^17/3^3
3^(2n+1)=3^(2×16)/3^3
3^(2n+1)=3^29
3^2n=3^28
2n=28
n=14

d) (6^n)^4×36=216^10
6^4n×6^2=6^(3×10)
6^(4n+2)=6^30
6^4n=6^28
4n=28
n=7

e)7^(n^2)÷7=49^24
7^(n^2-1)=7^(2×24)
7^(n^2)=7^49
n^2=49
n=7

g)15^(n+4)÷5^(n+4)=81^6
3^(n+4)×5^(n+4)÷5^(n+4)=3^(4×6)
3^(n+4)=3^24
n=20

h)81^n÷9^n+9^(n+2)÷9=90÷9^6
9^2n÷9^n+9^(n+2)÷9=9*10/9^6
9^n+9^(n+1)=10/9^5
I don't know where to go from here

I)what?
7 0
3 years ago
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