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3 years ago

Plzzzzzzzzz help fast

Mathematics

2 answers:

SCORPION-xisa [38]3 years ago

6 0

Answer:

can u write the question

Step-by-step explanation:

Marrrta [24]3 years ago

6 0

Answer:

a translation 5 units to the left

Step-by-step explanation:

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Use f(x) = 2x-9 and g(x)=9-x^2 to evaluate g(f(3))

Keith_Richards [23]

Answer:

g(f(3)) = 0

Step-by-step explanation:

f(x) = 2x - 9

f(3) = 2(3) - 9

f(3) = -3

g(-3) = 9 - (-3)^2

g(-3) = 9 - 9

g(-3) = 0

8 0

3 years ago

Find the equation of the plane that passes through the point (4,3,−2)(4,3,−2) and contains the line x(t)=−4−4t,y(t)=−1+3t,z(t)=1

aniked [119]

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3 years ago

For which equation would x = 5 be a solution?

Pachacha [2.7K]

3x+6=21
3x=21-6
3x=15
x=5

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3 years ago

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Can you divide an angle into three even smaller angles using any geometric tools?

Elanso [62]

An angle can be bisected using just a compass
and straight edge, but it can't be trisected.

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3 years ago

8. For many computer tablets, the owner can set a 4-digit pass code to lock the device.

Anton [14]

Answer:

a. There 5040 different pass codes if the digits cannot be repeated

b. The probability that the pass code is 1234 is ≈ 0.0002

c. The probability that two people both have a pass code of 1234 is 4× $10^{-8}$

Step-by-step explanation:

a. How many different 4-digit pass codes are possible if the digits cannot be repeated?

There are

10 possibilities for the first digit

9 possibilities for the second digit

8 possibilities for the third digit

7 possibilities for the fourth digit

Thus there are 10×9×8×7=5040 different pass codes

b. If the digits of a pass code are chosen at random and without replacement from the digits, what is the probability that the pass code is 1234

The probability that the first digit is 1 is $\frac{1}{10}$

The probability that the second digit is 2 is $\frac{1}{9}$

The probability that the second digit is 3 is $\frac{1}{8}$

The probability that the fourth digit is 4 is $\frac{1}{7}$

Thus the probability that the pass code is 1234 is $\frac{1}{10} * \frac{1}{9} *\frac{1}{8} *\frac{1}{7} =\frac{1}{5040}$ ≈ 0.0002

c. The probability that two people, who both chose a pass code by selecting digits at random and without replacement, both have a pass code of 1234?

The probability that the pass code of the one person is 1234 is 0.0002

The probability that the pass code of the other person is 1234 is 0.0002

Thus, the probability that two people have both pass code of 1234 is

0.0002×0.0002 = 4× $10^{-8}$

3 0

4 years ago