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valkas [14]
3 years ago
9

Jose ran 2.6 miles. Patel ran 2.60 miles. Who ran farther?

Mathematics
2 answers:
AfilCa [17]3 years ago
7 0

Answer:


Step-by-step explanation:

So they both ran the same distance because is you put 2.6 in money its 2.60 therefore their equal

Verizon [17]3 years ago
5 0

They ran the same amount of miles, because the zero at the end doesn't count.

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According to the property √a/√b=√a/b, which choice is equivalent to the quotient below? √35/√7​
Ilya [14]

Answer:

D

Step-by-step explanation:

\frac{\sqrt{35} }{\sqrt{7} } =\frac{\sqrt{5} \times \sqrt{7} }{\sqrt{7} } =\sqrt{5}

4 0
3 years ago
How do u work this out and what's the answer
I am Lyosha [343]
1) you would need to turn both fractions so they have a common denominator, that would be 2 2/4 and 3/4
2) then subtract : 2 2/4 - 3/4, and you would subtract numerators and the whole number, but you keep the denominator the same. However, you cannot subtract 2 and 3 in this case, so you need to change 2 2/4 to 1 5/4 (they are still equivalent)
3) 1 5/4 - 3/4 = 1 2/4, which simplified version is 1 1/2
Therefore, the answer is 1 and 1/2
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3 years ago
Mr. Bridges paid $582.40 for a set of four tires for a truck about how much was the cost per tire
skelet666 [1.2K]
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5 0
3 years ago
EFGH is translated 3 units to the left and 7 units up.<br> What are the coordinates of H'?
Nikitich [7]
Where was H originally?
8 0
3 years ago
Read 2 more answers
Hiiii.. please help me with this limit question ​
Alenkasestr [34]

Answer:

π

Step-by-step explanation:

Solving without L'Hopital's rule:

lim(x→0) sin(π cos²x) / x²

Use Pythagorean identity:

lim(x→0) sin(π (1 − sin²x)) / x²

lim(x→0) sin(π − π sin²x) / x²

Use angle difference formula:

lim(x→0) [ sin(π) cos(-π sin²x) − cos(π) sin(-π sin²x) ] / x²

lim(x→0) -sin(-π sin²x) / x²

Use angle reflection formula:

lim(x→0) sin(π sin²x) / x²

Now we multiply by π sin²x / π sin²x.

lim(x→0) [ sin(π sin²x) / x² ] (π sin²x / π sin²x)

lim(x→0) [ sin(π sin²x) / π sin²x] (π sin²x / x²)

lim(x→0) [ sin(π sin²x) / π sin²x] lim(x→0) (π sin²x / x²)

π lim(x→0) [ sin(π sin²x) / π sin²x] [lim(x→0) (sin x / x)]²

Use identity lim(u→0) (sin u / u) = 1.

π (1) (1)²

π

Solving with L'Hopital's rule:

If we plug in x = 0, the limit evaluates to 0/0.  So using L'Hopital's rule:

lim(x→0) [ cos(π cos²x) (-2π cos x sin x) ] / 2x

lim(x→0) [ -π cos(π cos²x) sin(2x) ] / 2x

-π/2 lim(x→0) [ cos(π cos²x) sin(2x) ] / x

Again, the limit evaluates to 0/0.  So using L'Hopital's rule one more time:

-π/2 lim(x→0) [ cos(π cos²x) (2 cos(2x)) + (-sin(π cos²x) (-2π cos x sin x)) sin(2x) ] / 1

-π/2 lim(x→0) [ 2 cos(π cos²x) cos(2x) + π sin(π cos²x) sin²(2x) ]

-π/2 (-2)

π

8 0
3 years ago
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