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Levart [38]
3 years ago
14

HIIII

Mathematics
2 answers:
kicyunya [14]3 years ago
7 0

Answer:

3.21

Step-by-step explanation:

got it from googIe

VikaD [51]3 years ago
5 0

Answer:

I got 3.21 by subtracting them

10.00

-6.79

____

3.21

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Answer for Answer? I need the answer to 8 and 9. I’ll answer 2 of yours
steposvetlana [31]
Let's go:

8)

Total of senior students = 38
Total of senior students that preferred evening classes: 7
7/38 = aproximately 0,18 or 18%

Alternative A)

9)
Imagine being sold 100 desserts. Of these 100 desserts, 75% até chocolate, meaning 75 of these desserts are chocolate. Of these 75, 23% contains nuts. So:

75 × 0,23 = 17,25 = aproximately 17,3%

Alternative C)

I hope I helped you.
6 0
3 years ago
Read 2 more answers
one more than six times that number is 25. select the solution and graph that represents the original number
Brums [2.3K]

6n+1 = 25

subtract 1 from each side

6n+1-1 = 25-1

6n = 24

divide each side by 6

6n/6 = 24/6

n =4


6 0
3 years ago
Kyle buys 3 bags of Takis at the store. He pays $4.50 to the cashier. How much does each bag of Takis cost?
trapecia [35]

Answer:

1.50$

Step-by-step explanation:

4.50/3= 1.50

Hope this helps.

8 0
3 years ago
Read 2 more answers
Please math help<br> X/9+8=14
Novosadov [1.4K]
X/9+8=14

First subtract the 8 from 14
14-8
x/9=6

Next multiply both sides by 9

6*9=54

x=54
8 0
3 years ago
Read 2 more answers
Let R be the region bounded by
loris [4]

a. The area of R is given by the integral

\displaystyle \int_1^2 (x + 6) - 7\sin\left(\dfrac{\pi x}2\right) \, dx + \int_2^{22/7} (x+6) - 7(x-2)^2 \, dx \approx 9.36

b. Use the shell method. Revolving R about the x-axis generates shells with height h=x+6-7\sin\left(\frac{\pi x}2\right) when 1\le x\le 2, and h=x+6-7(x-2)^2 when 2\le x\le\frac{22}7. With radius r=x, each shell of thickness \Delta x contributes a volume of 2\pi r h \Delta x, so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\dfrac{\pi x}2\right)\right) \, dx + 2\pi \int_2^{22/7} x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56

c. Use the washer method. Revolving R about the y-axis generates washers with outer radius r_{\rm out} = x+6, and inner radius r_{\rm in}=7\sin\left(\frac{\pi x}2\right) if 1\le x\le2 or r_{\rm in} = 7(x-2)^2 if 2\le x\le\frac{22}7. With thickness \Delta x, each washer has volume \pi (r_{\rm out}^2 - r_{\rm in}^2) \Delta x. As more and thinner washers get involved, the total volume converges to

\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^{22/7} (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16<em />

d. The side length of each square cross section is s=x+6 - 7\sin\left(\frac{\pi x}2\right) when 1\le x\le2, and s=x+6-7(x-2)^2 when 2\le x\le\frac{22}7. With thickness \Delta x, each cross section contributes a volume of s^2 \Delta x. More and thinner sections lead to a total volume of

\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^{22/7} \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70

7 0
1 year ago
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