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3 years ago

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

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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

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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