Question

two rain barrels are used to water a school garden. the table and equation represent the amount of water remaining in each barrel, y, in gallons, x minutes since the students began watering the garden.
barrel 1
barrel 2

x	y
1	50
3	40
5	30
7	20

$y = -5x + 60$
which statement is true?
a barrel 1 initially contains more water than barrel 2.
b barrel 1 drains water at a faster rate than barrel 2.
c barrel 1 and barrel 2 initially contain the same amount of water.
d barrel 1 and barrel 2 drain water at the same rate.
thursday
what value of y satisfies the equation $\frac{2}{3}(y + 6) = \frac{1}{3}y + 7$?
$6.27 \times 10^{-7} + 5.81 \times 10^{-9}$
$(2^3 \times 2^{-6})^2$
$2^{2} \times 2^{-12} = 2^{-10}$

Explanation:

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(Rain Barrel Problem)

Step1: Find Barrel 1's rate

Rate = $\frac{40-50}{3-1} = \frac{-10}{2} = -5$ gallons/min

Step2: Find Barrel 1's initial water

Use $y=mx+b$, $50=-5(1)+b$ → $b=55$ gallons

Step3: Compare to Barrel 2

Barrel2: $y=-5x+60$, rate=-5, initial=60 gallons

Step4: Evaluate options

A: 55 < 60 → False; B: Rates equal → False; C: 55≠60 → False; D: Rates both -5 → True
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(Algebra Equation Problem)

Step1: Eliminate denominators

Multiply all terms by 3: $2(y+6) = y + 21$

Step2: Expand left side

$2y + 12 = y + 21$

Step3: Isolate y terms

$2y - y = 21 - 12$

Step4: Solve for y

$y=9$
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(Scientific Notation Addition)

Step1: Rewrite terms to same exponent

$6.27 \times 10^{-7} = 627 \times 10^{-9}$

Step2: Add the terms

$627 \times 10^{-9} + 5.81 \times 10^{-9} = (627+5.81) \times 10^{-9} = 632.81 \times 10^{-9}$

Step3: Convert to proper notation

$6.3281 \times 10^{-7}$
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(Exponent Simplification)

Step1: Add exponents inside parentheses

$2^{3 + (-6)} = 2^{-3}$

Step2: Apply outer exponent

$(2^{-3})^2 = 2^{-3 \times 2} = 2^{-6}$

Answer:

D. Barrel 1 and Barrel 2 drain water at the same rate.
$y=9$
$6.3281 \times 10^{-7}$
$2^{-6}$