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2 years ago

If a randomizer picks seven letters from the alphabet (with repeats allowed), how many possible outcomes are there?

Mathematics

1 answer:

Virty [35]2 years ago

7 0

Answer:

7×26=182

Step-by-step explanation:

I'm pretty sure that's the right answer but I just multiplied 26 by 7. Hope that helped!

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Solve for x: 1+4x=13+x

topjm [15]

ANSWER
x = 4

EXPLANATION
1) 1+4x = 13+x

2) 1+3x = 13

3) 13 - 1 = 12

4) 3x = 12 (Solve for x by dividing 3 on both sides)

5) x = 4

7 0

2 years ago

-3/2 multiplied by 3?

rosijanka [135]

Let's solve this step-by-step! :)

-(3/2) * 3

= -(9/2)

I hope this helped!! :D

7 0

3 years ago

Read 2 more answers

Please tell me the steps to this problem. i will give brainliest

gayaneshka [121]

Option C: -3 is the average rate of change between $x=-1$ and $x=1$

Explanation:

The formula to determine the average rate of change is given by

Average rate of change = $\frac{y_2-y_1}{x_2-x_1}$

We need to find the average rate of change between $x=-1$ and $x=1$

Thus, from the table, we have,

$x_1=-1$ , $y_1=5$

$x_2=1$ , $y_2=-1$

Thus, substituting these values in the formula, we get,

Average rate of change = $\frac{-1-5}{1+1}$

$=\frac{-6}{2}$

$=-3$

Thus, the average rate of change between $x=-1$ and $x=1$ is -3.

Hence, Option C is the correct answer.

6 0

2 years ago

T(x) = -2(-x - 3)' + 1

Anna [14]

Answer:

Step-by-step explanation:

3 0

2 years ago

Find a and b such that gcd(a, b) = 14, a > 2000, b > 2000, and the only prime divisors of a and b are 2 and 7.

choli [55]

Answer:

$a = 2^\alpha.14 \ ,\alpha\geq 8\\b = 7^\delta.14 \ , \delta\geq 3$

Step-by-step explanation:

As 2 and 7 are the only prime divisors of both $a$ and $b$ we know that both can be written as:

$a = 2^\alpha . 7^\beta \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\$

Where $\mathbb{N}_{0}$ is the set of all natural numbers adding the zero (careful because this part is important as I'll explain next).

We also know that 14 divides both numbers and that is actually the greatest common divisor between them. So we can rewrite a and b as follows:

$a = 2^\alpha . 7^\beta.14 \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta.14 \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\$

Why do I write them like this? Because this way is easier to observe that if $\alpha$ and $\gamma$ were both greater than zero, then 28 would divide both hence 14 wouldn't be their g.c.d.. Likewise, if $\beta$ and $\delta$ were both greater than zero, then 98 would divide both and once again, 14 wouldn't be their g.c.d.

So either of them has to be equal to zero. And then we have that

$a = 2^\alpha.14 \\b = 7^\delta.14$

All we have left to do is find the possible values for $\alpha$ and $\delta$ so that $a>2000 \ , b>2000$ and that only happens if $\alpha\geq 8$ and $\delta\geq 3$

8 0

3 years ago