Mathematics, 10.12.2019 03:31 hrcatena

a fuel tank contains 1 3/4 gallons of gasoline. casey adds 33 1/3 gallons of
gasoline to the tank. how many gallons of gas does the tank now contain?
gallons
add mixed numbers with unlike denominators word problems

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Answer from: lanc3r3vo

The gallons of gas the tank contain now is 35 1/12.

Step-by-step explanation:

Given:

Fuel tank contains 1 3/4 gallons of gasoline.

Casey adds 33 1/3 gallons of gasoline to the tank.

Now, to find the total gallons of gas the tank contain.

Converting the mixed fractions into improper.

Gallons of gasoline in fuel tank $=1\frac{3}{4}=\frac{7}{4}$ .

Gallons of gasoline Casey adds $=33\frac{1}{3} =\frac{100}{3}$ .

According to question:

By adding we get the total gallons of gasoline now tank contain:

$\frac{7}{4} +\frac{100}{3}$

$=\frac{7\times 3+100\times 4}{12}$

$=\frac{21+400}{12}$

$=\frac{421}{12}$

$=35\frac{1}{12}$

Therefore, the gallons of gas the tank contain now is 35 1/12.

Answer from: Quest

5. y=6; de=7; fg=8

6. x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

y = 6

of course, the total chord length is the sum of the lengths of its segments.

de = dh +eh = 3+4 = 7

fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

x = lk = mk×nk/jk = 6×7/10 = 4.2

jl = jk +lk = 10 +4.2 = 14.2

mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.