16/8 into a mixed number

Answers

Answer 1

Answer: ok well the answer will be 2 because 8+8=16 and 8 can go into 16 2 times.

Answer 2

Answer: the answer is
2/1
because 8 times 2 is 16 and if you divide 16 by 8 u will get 2 and 8 divided by 8 is 1

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X+y=1 show work and graph it please. and use x/y.

Answers

you will put x=0 and calculate the value of y=1 and then put y=0 and get value of x=1 and then put x=1 and get value of y=0 and then put y=1 and get x and so one , finally you could plot the graph.

10 1/2 cups of flour are needed to make 3 loaves of bread. How many cups are needed for 12?

Answers

Answer:

42 cups of flour

Step-by-step explanation:

12 is 4 times more than 3 (3 × 4 = 12), so we know that we need 4 times 10 1/2 (equal to 10.5) cups of flour.

All we need to do for this is do 10.5 × 4 = x and solve for x.

10.5 × 4 = 42 cups of flour

Solve for b if 7+16b=29+5b

Answers

7 + 16b = 29 + 5b

Keep the variables on one side

16b - 5b = 29 - 7

11b = 22

Divide by 11 isolate b:

$(11b)/(11)$ = $(22)/(11)$
11 and 11 cancels out

b = 2

check:
7 + 16(2) = 29 + 5(2)
7 + 32 = 29 + 10
39 = 39

There are 291 calories in three ounces of a certain ice cream. How many calories are there in two pounds?

Answers

3,436 to 3,752 calories in 2 pounds

Whats 3x+27=39 will mark brainliest

Answers

Answer:

x=4

Step-by-step explanation:

Answer:

Step-by-step explanation:

5x+39=3x+27 i think your welcome :)

I am writing this question only to see how smart the community on Brainly is.What is the 3rd derivative of $f(x)=sin(x)x^3+x^2tan(x)$

Answers

I am very confused as to why you posed this question in the way you did. People would have answered your question regardless.

First derivative:
f'(x) = 3x^2sin(x) + x^3cos(x) + 2xtan(x) + x^2sec^2(x)

Second derivative:
f''(x) = 6xsin(x) + 3x^2cos(X) + 3x^2cos(x) - x^3sin(x) + 2tan(x) + 2xsec^2(x) + 2xsec^2(x) + 2x^2sec^2(x)tan(x)

f''(x) = 6xsin(x) + 6x^2cos(x) - x^3sin(x) + 2tan(x) + 4xsec^2(x) +

Third Derivative:
f'''(x) = 6sin(x) + 6xcos(x) + 12xcos(x) - 6x^2sin(x) - 3x^2sin(x) - x^3cos(x) + 2sec^2(x) + 8xsec^2(x)tan(x) + 4sec^2(x) + 4x^2sec^2(x)tan^2(x) + 4xsec^2(x)tan(x) + 2x^2sec^4(x)

f'''(x) = 6sin(x) + 18xcos(x) - 9x^2sin(x) - x^3cos(x) + 12xsec^2(x)tan(x) + 6sec^2(x) + 4x^2sec^2(x)tan^2(x) + 2x^2sec^4(x)

f'''(x) = (6 - 9x^2)sin(x) + (18x - x^3)cos(x) + 12xsec^2(x)tan(x) + 6sec^2(x) + 4x^2sec^2(x)tan^2(x) + 2x^2sec^4(x)

There are many more ways this can be factored, but if this is just to test whether someone knows the basics of calculus, this should be sufficient.

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