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Firlakuza [10]
2 years ago
15

Joe gave the following argument: Since lim x→0 0 = 0,

Mathematics
1 answer:
lions [1.4K]2 years ago
7 0
No error - A and B are completely right, and using the subtraction law D is right
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aleksklad [387]
Turn the triangle around, because you have to reflect it, if you need extra help lmk
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3 years ago
HELP PLEASE I NEED IT ASAP
Akimi4 [234]

Answer:

849.9 cm squared

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Can someone help me to answer this please.
Volgvan

All it's asking is:  How many cubic inches are there in 22 gallons ? And it already GAVE us the conversion factor !

The conversion factor is . . . . .  1 gallon = 231 cubic inches

Multiply each side by 22:  22 gallons = 22 x (231 cubic inches)

22 x 231 = 5,082

22 gallons = 5,082 cubic inches

If you empty a full 22-gallon tank out through that flow meter, the meter  will register 5,082 cubic inches when the tank is empty.


6 0
3 years ago
There are 4,389 dogwood trees in the state park. The park workers are going to plant 342 more trees. How many trees will there b
Dmitrij [34]

Answer:

4731

Step-by-step explanation:

4,389+342=4,731

6 0
3 years ago
Is sin theta=5/6, what are the values of cos theta and tan theta?
mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill

\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}

5 0
3 years ago
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