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lord [1]
3 years ago
6

he access code for a​ car's security system consists of five digits. The first digit cannot be 2 and the last digit must be odd.

How many different codes are​ available? he access code for a​ car's security system consists of five digits. The first digit cannot be 2 and the last digit must be odd. How many different codes are​ available?
Mathematics
1 answer:
SpyIntel [72]3 years ago
7 0
Using numbers 0-9 ( 10 numbers for each digit)
 1st digit can't be 2 so it can be 10-1 = 9 other digits
2nd, 3rd and 4th digits can be 1 of 10 digits each

5th digit must be odd do it can be 1, 3, 5, 7, 9 ( 1 of 5 digits)

so number of different codes would be 9 x 10 x 10 x 10 x 5  = 45,000 different codes

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Prove the identity 2csc2x=csc^2xtanz
Verizon [17]

Step-by-step explanation:

Consider the LHS, after the 5th step, consider the RHS

2 \csc(2x)  =  \csc {}^{2} (x)  \tan(x)

2 \frac{1}{ \sin(2x) }  =  \csc {}^{2} (x)  \tan(x)

2  \times \frac{1}{2 \sin(x)  \cos(x) }  =  \csc {}^{2} (x)  \tan(x)

\frac{1}{ \sin(x) \cos(x)  }  =   \csc {}^{2} (x)  \tan(x)

\csc(x)  \sec(x)  =  \csc {}^{2} (x)  \tan(x)

Consider the RHS

\csc(x)  \sec(x)  =( 1 +  \cot {}^{2} (x) ( \tan(x))

\csc(x)  \sec(x)  =  \tan(x)  +  \cot(x)

\csc(x)  \sec(x)  =  \frac{ \sin(x) }{ \cos(x) }  +  \frac{ \cos(x) }{ \sin(x) }

\csc(x)  \sec(x)  =  \frac{1}{ \sin(x) \cos(x)  }

\csc(x)  \sec(x)  =  \csc(x)  \sec(x)

7 0
2 years ago
2. What is A-2) if f(x)= ?<br> F.-2<br> G. 1<br> H. O<br> I. 1
NARA [144]

Answer:

F.-2...........................................

8 0
4 years ago
X+3=12<br> show steps<br> thx
AlekseyPX

Answer:

are u serious this is the most easiest question ive ever answered but the answer is 9 because u can minus  12-3 and get 9 so x=9

Step-by-step explanation:

4 0
3 years ago
2 cans of paint hold 64 and 125 oz. if the cans are similar cylinders and the smaller lid has an area of 18 find the area of the
kramer

Answer:

The area of the bigger lid is approximately 29.2 square inches

Step-by-step explanation:

The given data are;

The volume held by one of the cans of paint, V₁ = 64 oz

The volume held by the other cans of paint, V₂ = 125 oz

The area of the smaller lid = 18 in²

Therefore,

The 64 Oz = The smaller can of paint

The 125 oz = The larger can of paint

The diameter of the smaller lid, d₁ = √(18 × 4/π) ≈ 4.787

The ratio of can volumes = (The ratio of can diameters)³

V₁/V₂ = (d₁/d₂)³

V₁/V₂ = 64/125

d₁/d₂ ≈ 4.787/d₂

64/125 = (4.787/d₂)³

4.787/d₂ = ∛(64/125) = 4/5

d₂ ≈ 4.878×5/4 = 6.0975

The area of the bigger lid, A₂ = π·d₂²/4

∴ A₂ = π × 6.0975²/4 ≈ 29.2

The area of the bigger lid, A₂ ≈ 29.2 in.²

6 0
3 years ago
Hope you can help:)
DanielleElmas [232]
Through some guessing I concluded it is C, but I'm not completely sure, either that or B<span />
4 0
3 years ago
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