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shusha [124]
3 years ago
11

Helop me plsineed an aswer​

Mathematics
1 answer:
Andrew [12]3 years ago
4 0
A: ❤️❤️/❤️❤️❤️

B: ❤️❤️❤️❤️❤️❤️❤️/❤️❤️❤️❤️❤️❤️❤️❤️❤️

C: ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️/❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️

D: ❤️❤️❤️❤️❤️❤️❤️❤️/❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️

E: ❤️❤️❤️❤️/❤️❤️❤️❤️❤️

F: ❤️❤️❤️❤️❤️❤️❤️❤️/❤️❤️❤️❤️❤️❤️

G: ❤️❤️❤️❤️❤️❤️❤️❤️❤️/❤️❤️❤️❤️

H: ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️
❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️
❤️❤️❤️❤️❤️❤️❤️❤️/❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️
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Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating
MakcuM [25]

Answer:

A=8.4063u^{2}

Step-by-step explanation:

Be the functions:

y=\frac{3}{x};y=\frac{3}{x^{2}}:x=7

according the graph:

\int\limits^1_7 {\frac{3}{x} } \, dx -\int\limits^1_7 {\frac{3}{x^{2} } } \, dx =3\int\limits^1_7 {\frac{1}{x} } \, dx -3\int\limits^1_7 {\frac{1}{x^{2} } } \, dx=3(\int\limits^1_7 {\frac{1}{x} } \, dx -\int\limits^1_7 {\frac{1}{x^{2} } } \, dx)=3[lnx-\frac{1}{x}](1-7)=3[(ln7-ln1)-(\frac{1}{7}-1)]=3[(1.945-0)-(0.1428-1)]=3*(1.945+0.8571)=3*2.8021=8.4063u^{2}

6 0
4 years ago
Please help on this math homework !
Savatey [412]

Answer:It's c

I already did it lesson

6 0
4 years ago
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Evaluate when x = 3, y = 2, and z = 4:
Lelu [443]

\huge\text{Hey there!}

\large\text{If x = 3, y = 2, and z = 4 then substitute it in the given equation!}

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A student has selected 8 books that she likes, but she has money only for 3 books. how many possible selections does she have?
ladessa [460]
Since the order of the books chosen doesn't matter, we use the combination formula to find the number of possible selections. Recall that a combination of k items from a set of n items is given by   nCk= \frac{n!}{k!(n-k)!}. 

In this problem, n = 8 and k = 3. So the number of possible selections is 
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3 years ago
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The temperature T in degrees Fahrenheit is (approximately) a linear function of the number C of cricket chirps per minute. Twent
Law Incorporation [45]

Answer:

a) T= 0.25 C+37

b) T= 62°F

Step-by-step explanation:

Here T is a linear function of C

T= mC+b

For each additional chirp C corresponding to an increase in T of 0.25

therefore, m= 0.25

T= 42° F   C=20 ( given)

therefore,

a) 42= 0.25×20+b

b= 37

therefore T= 0.25 C+37

b) now C= 100

therefore, T= 0.25×100+37 = 62°F

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