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ohaa [14]
3 years ago
7

What is the value of x in the solution of the following system of equations: 3x + 2y = 12 and 5x - 2y = 4 ?

Mathematics
2 answers:
Dima020 [189]3 years ago
4 0

Answer:B

Step-by-step explanation:

stellarik [79]3 years ago
4 0

Answer: x=  4  −  2 y /3

Step-by-step explanation:

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Plz help me<br><br>Use the table chart and find out what the alitude of 5,000 feet.
mash [69]
The is (5000,59) and you can check this by doing this -0.004(5000)+79
6 0
3 years ago
Frank has a 2-digit number on his baseball uniform. the number is a multiple of 10 and has 3 for one its factors. what three num
Feliz [49]
A. What three numbers could possibly be on his uniform.
B. Probability of being a multiple of 10 and is two digits and has a 3 as a factor.
C. Frank has a 30 on his uniform. The number is a multiple of 10 and one factor of the number is 3.
4 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
I need help! It's timed. Input - Output is the picture that is linked.
KonstantinChe [14]

Answer:

5

Step-by-step explanation:

substitute 2 for t

-16(2)^2+32(2)+5=5

6 0
3 years ago
HELPP ME PLSSSSSSS I NEED IT NOW PLSSS. How could you calculate the x-coordinate of the midpoint of a horizontal
Romashka [77]
C? i’m pretty sure sorry if it’s wrong
7 0
3 years ago
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