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2 years ago

Jeremy used 11 gallons and 3 quarts of water to water his garden. How many quarts of water did Jeremy use?

Mathematics

1 answer:

Anon25 [30]2 years ago

6 0

Answer:

Step-by-step explanation:

since each gallon is 4 quarts you would multiply by 11. 44 quarts then add 3 extra quarts to get 47

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William has ten colored game pieces in a box: 3 yellow pieces, 5 orange pieces, 1 green piece, and 1 red piece. What is the prob

Nutka1998 [239]

Answer:

Step-by-step explanation:

yellow= .30

orange= .50

green= .10

red- .10

6 0

3 years ago

suppose that the local sales tax rate is 6.25% and youpurchased a used carr for 16,800. what is the car's total cost

Gwar [14]

Total= Car Cost + (Car Cost * Tax %)

Total= $16,800 + ($16,800 * 6.25%)

convert % to decimal form; divide % by 100

Total= $16,800 + ($16,800 * 0.0625)

multiply inside parentheses

Total= $16,800 + $1,050

Total= $17,850

ANSWER: The total cost for the used car with tax is $17,850.

Hope this helps! :)

8 0

3 years ago

What is the d/dx tan^3x

Bad White [126]

$\frac{d}{dx}(tan^{3}x) = \frac{d}{dx}(tanx)^{3}$
$= 3(tanx)^{2} \cdot sec^{2}x$

$\therefore \frac{d}{dx}(tan^{3}x) = 3tan^{2}x \cdot sec^{2}x$

6 0

3 years ago

If bob has five apples and he eats 2 how many does he have left

qwelly [4]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

If g (x) = 1/x then [g (x+h) - g (x)] /h

lys-0071 [83]

Answer:

$\dfrac{-1}{x(x+h)}, h\ne 0$

Step-by-step explanation:

If $g(x) = \dfrac{1}{x}$ , then $g(x+h) = \dfrac{1}{x+h}$ . It follows that

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}$

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}$

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

$= \dfrac{-1}{x(x+h)}, h\ne 0$

5 0

3 years ago