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3 years ago

Gallon of gas: $2.15, 15% markup

Mathematics

1 answer:

Irina18 [472]3 years ago

3 0

The answer would be $2.47

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4. The figure shows a rectangle with a width of 6k inches. The length of the rectangle is 4 inches shorter than the width.

vesna_86 [32]

Answer:

Step-by-step explanation:

sorry if its not correct but i hope it helps

8 0

3 years ago

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10 lb 6oz minus 7lb 11oz is

Leya [2.2K]

(10 lb 6 oz)-(7 lb 11 oz) = 2.6875 lb

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3 years ago

Which equation represents the graph. Please help I’m trying to finish a post test for math

8090 [49]

Answer:

5(x+3) is your answer

Step-by-step explanation:

3 0

3 years ago

Write a function number_of_pennies() that returns the total number of pennies given a number of dollars and (optionally) a numbe

seraphim [82]

Ok, first of all, it is important to understand that this function would depend on 2 variables; the number of dollars (x) and the number of pennies (y). Every dollar is equivalent to 100 pennies. So, if we are given x dollars, we have 100*x pennies. If we also have y pennies, we still get y pennies. We need to add these two to get the total amount of pennies. Thus, the correct function is:
f(x,y)=100*x+y

7 0

3 years ago

Integrate 1 - x / x(x2 + 1) d x by partial fractions.

solniwko [45]

Answer:

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

Step-by-step explanation:

step 1:- by using partial fractions

$[tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }$ ......(1)

step 2:-

solving on both sides

$1-x=A(x^{2} +1)+(Bx+C)x......(2)$

substitute x =0 value in equation (2)

1=A(1)+0

A=1

comparing x^2 co-efficient on both sides (in equation 2)

0 = A+B

0 = 1+B

B=-1

comparing x co-efficient on both sides (in equation 2)

-1 = C

step 3:-

substitute A,B,C values in equation (1)

now

$\\\int\limits^ {} \, \frac{1-x}{x(x^{2}+1) } d x =\int\limits^ {} \frac{1}{x} d x +\int\limits^ {} \frac{-x}{x^{2}+1 } d x -\int\limits \frac{1}{x^{2}+1 } d x$

by using integration formulas

i) by using $\int\limits \frac{1}{x} d x =log x+c........(a)\\\int\limits \frac{f^{1}(x) }{f(x)} d x= log(f(x)+c\\$ .....(b)

$\int\limits tan^{-1}x dx =\frac{1}{1+x^{2} } +C$ .....(c)

step 4:-

by using above integration formulas (a,b,and c)

we get answer is

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

6 0

3 years ago