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dezoksy [38]
4 years ago
6

Miguel, Dean, and Joshua were candidates in an election in which 1,000 people voted. Joshua won the election, receiving 95 votes

more than Dean and 186 votes more than Miguel. How many votes did Dean receive?
Mathematics
1 answer:
gtnhenbr [62]4 years ago
4 0

Answer:jg jg

Step-by-step explanation:

ytffs

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Hello! Please answer (c). I'll put 16 points for this answer! I'm in middle school, not college. Please reply and take your time
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G= (-1,3)

H= (-4,0)

I= (3,-2)

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Can someone please answer this question?
Vlad [161]

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I believe the answer is A or D

Step-by-step explanation:

8 0
3 years ago
GIVING BRAINLEST TO THE PERSON THAT CAN EXPLAIN HOW TO DO THIS THE BEST :)
WINSTONCH [101]

Answer:

530 m²

Step-by-step explanation:

First of all we need to label the missing sides ,please check the figure I provided for that :

x = 37 - (10+ 15) = 37 - 25 = 12

z = 10

y = 2 × 10 = 20

The shade does area = area of the whole rectangle − the white area

                                    = 37 × 20 − [ (15×20)÷2 + (10×12)÷2 ]

                                    =  740      − [    150        +      60       ]

                                    = 740.      −               210

                                    = 530

5 0
3 years ago
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
2 years ago
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