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Mumz [18]
3 years ago
14

Evaluate the expression 8a+11−3b for a = 4 and b = 2.

Mathematics
1 answer:
Dvinal [7]3 years ago
7 0

Answer: 37

8(4)+11-3(2)

32+11-6

43-6

37

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Solve for the variable
liraira [26]

Answer:

y = 25

Step-by-step explanation:

30=\frac{6}{5} y\\\\\frac{6}{5} y=30\\\\(\frac{6}{5} y)5=30*5\\\\6y=150\\\\\frac{6y=150}{6}\\\\\boxed{y=25}

Hope this helps!

8 0
3 years ago
I cannot understand this can you guys help
igor_vitrenko [27]
The mean is not the same thing as median
mean=78+81+85+87=331
mean=331/4=82.75
median is the middle number, but you have a par set of numbers, so your median will be the middle numbers 78,81,85,87 , which are 81+85=166, 166/2=83
83 is your median
5 0
3 years ago
A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.
irakobra [83]

Answer:

The integral is equal to 5\sec^2(2x)+C for an arbitrary constant C.

Step-by-step explanation:

a) If u=\tan(2x) then du=2\sec^2(2x)dx so the integral becomes \int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C. (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case u=\sec(2x) hence du=2\tan(2x)\sec(2x)dx. We rewrite the integral as \int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C.

c) We use the trigonometric identity \tan(2x)^2+1=\sec(2x)^2 is part b). The value of the integral is 5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C. which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.  

3 0
3 years ago
How is circumference and area the same?
V125BC [204]
The area<span> and the </span>circumference<span> of a circle. Make sure that you remember that a</span>circumference<span> is a length and so is measured , , etc. Remember as well that </span>area<span> is measured in square units, , etc. All of these calculations will involve the use of Pi ( ) which you can take as being equal to .</span>
4 0
3 years ago
Read 2 more answers
Can you brake this down 455÷7​
Hatshy [7]
Its very easy
455 means 5*7*13

now divide (5*7*13) by 7

you will easily get 5*13=65
5 0
3 years ago
Read 2 more answers
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