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WITCHER [35]
3 years ago
14

A bike lock has a 3 digit combination. Each character can be any digit from 3-8. The only restriction is that all 3 characters c

annot be the same (e.G.555,666,777...). How many combinations are possible?
Mathematics
1 answer:
lesya [120]3 years ago
7 0

Answer:

The number of possible combinations is 180

Step-by-step explanation:

The numbers to select from are 3-8

These are 6 numbers

Now, the restriction we have is that the 3 characters cannot be the same

For the first character, we have 6 choices;

For the second character, we have another 6 choices

For the last character, since it cannot be the same, we can only have 5 choices

So the possible number of choices will be;

6 * 6 * 5 = 189 combinations

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What is true about the dilation? A large parallelogram is the pre-image and a smaller parallelogram is the image. It has a scale
kotegsom [21]

Answer:

- It is a reduction with a scale factor between 0 and 1.

- It is an enlargement with a scale factor greater than 1.

Step-by-step explanation:

1. Dilation is defined as a transformation of an image in which the size on this image changes but its shape does not change.

2. In dilation, to obtain the new dimensions of the image you must multiply the dimensions of the original image by a number called "Scale factor".

3. When the scale factor is greater than 1 it is an enlargement and when this is between 0 and 1, it is a reduction.

8 0
2 years ago
Read 2 more answers
What is the ratio between the LCM of 180 and 594 and the GCF of 180 and 594
aalyn [17]

180:2=90\\90:2=45\\45:3=15\\15:3=5\\5:5=1\\\\180=\boxed{2}\cdot2\cdot\boxed{3}\cdot\boxed{3}\cdot5\\\\594:2=297\\297:3=99\\99:3=3\\33:3=11\\11:11=1\\\\594=\boxed{2}\cdot\boxed{3}\cdot\boxed{3}\cdot3\cdot11

LCM(180,\ 594)=\boxed{2}\cdot\boxed{3}\cdot\boxed{3}\cdot2\cdot3\cdot5\cdot11=5940\\\\GCF(180,\ 594)=\boxed{2}\cdot\boxed{3}\cdot\boxed{3}=18

5 0
3 years ago
Find the sum for each and present - 66 + 42​
Fed [463]

-24

-66+42

=-24 since it is negative

3 0
3 years ago
Read 2 more answers
For the love of God help me !! I'm desperate for it tomorrow
Eduardwww [97]
Try to relax.  Your desperation has surely progressed to the point where
you're unable to think clearly, and to agonize over it any further would only
cause you more pain and frustration.
I've never seen this kind of problem before.  But I arrived here in a calm state,
having just finished my dinner and spent a few minutes rubbing my dogs, and
I believe I've been able to crack the case.

Consider this:  (2)^a negative power = (1/2)^the same power but positive.

So: 
Whatever power (2) must be raised to, in order to reach some number 'N',
the same number 'N' can be reached by raising (1/2) to the same power
but negative.

What I just said in that paragraph was:  log₂ of(N) = <em>- </em>log(base 1/2) of (N) .
I think that's the big breakthrough here.
The rest is just turning the crank.

Now let's look at the problem:

log₂(x-1) + log(base 1/2) (x-2) = log₂(x)

Subtract  log₂(x)  from each side: 

log₂(x-1) - log₂(x) + log(base 1/2) (x-2) = 0

Subtract  log(base 1/2) (x-2)  from each side:

log₂(x-1) - log₂(x)  =  - log(base 1/2) (x-2)  Notice the negative on the right.

The left side is the same as  log₂[ (x-1)/x  ]

==> The right side is the same as  +log₂(x-2)

Now you have:  log₂[ (x-1)/x  ]  =  +log₂(x-2)

And that ugly [ log to the base of 1/2 ] is gone.

Take the antilog of each side:

(x-1)/x = x-2

Multiply each side by 'x' :  x - 1 = x² - 2x

Subtract (x-1) from each side:

x² - 2x - (x-1) = 0

x² - 3x + 1 = 0

Using the quadratic equation, the solutions to that are
x = 2.618
and
x = 0.382 .

I think you have to say that <em>x=2.618</em> is the solution to the original
log problem, and 0.382 has to be discarded, because there's an
(x-2) in the original problem, and (0.382 - 2) is negative, and
there's no such thing as the log of a negative number.


There,now.  Doesn't that feel better. 
 






4 0
3 years ago
Find a polynomial of degree 3 with real coefficients and zeros of -3,-1, and 4, for which f(-2)=-24
Zanzabum

We want to find a polynomial

<em>f(x)</em> = <em>a</em> <em>x</em>³ + <em>b</em> <em>x</em>² + <em>c</em> <em>x</em> + <em>d</em>

such that the roots of <em>f</em> are <em>x</em> = -3, <em>x</em> = -1, and <em>x</em> = 4, and <em>f(x)</em> takes on a value of -24 when <em>x</em> = -2.

The factor theorem for polynomials tells us that we can factorize <em>f(x)</em> as

<em>a</em> <em>x</em>³ + <em>b</em> <em>x</em>² + <em>c</em> <em>x</em> + <em>d</em> = <em>a</em> (<em>x</em> + 3) (<em>x</em> + 1) (<em>x</em> - 4)

Expand the right side:

(<em>x</em> + 3) (<em>x</em> + 1) (<em>x</em> - 4) = <em>x</em>³ - 13<em>x</em> - 12

So we have

<em>a</em> <em>x</em>³ + <em>b</em> <em>x</em>² + <em>c</em> <em>x</em> + <em>d</em> = <em>a x</em>³ - 13<em>a</em> <em>x</em> - 12<em>a</em>

<em />

In order for both sides to be equal, the coefficients of both polynomials on terms of the same degree must be equal. This means

<em>a</em> = <em>a</em> (of course)

<em>b</em> = 0 (there is no <em>x</em>² term on the right)

<em>c</em> = -13<em>a</em>

<em>d</em> = -12<em>a</em>

<em />

We also have that <em>f</em> (-2) = -24, which means

<em>f</em> (-2) = <em>a</em> (-2 + 3) (-2 + 1) (-2 - 4)

-24 = 6<em>a</em>

<em>a</em> = -4

which in turn tells us that <em>c</em> = 52 and <em>d</em> = 48.

So we found

<em>f(x)</em> = -4<em>x</em>³ + 52<em>x</em> + 48

4 0
2 years ago
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