Helppp pleaseee

Consider generating length-7 strings of lowercase letters. How many strings are there that either begin with 2 consonants or end

Semmy [17]

Answer:

5259544316

Step-by-step explanation:

Given that:

Length of string = 7

Either begins with 2 consonants or ends with 2 vowels :

Either or :

A U B = A + B - (AnB)

Number of vowels in alphabet = 5

Number of consonants = 21

2 consonants at beginning :

First 2 consonants, then the rest could be any:

21 * 21 * 26 * 26 * 26 * 26 * 26 = 5239686816

3 vowels at the end :

First 4 letters could be any alphabet ; last 3 should be vowels.:

26 * 26 * 26 * 26 * 5 * 5 * 5 = 57122000

2 consonants at beginning and 3 vowels at the end :

21 * 21 * 26 *26 *5* 5 * 5 = 37264500

Hence,

2 consonants at beginning + 3 vowels at end 2 consonants at beginning - 2 consonants at beginning and 3 vowels At end

(5239686816 + 57122000) - 37264500

= 5259544316

Hence, number of 7 alphabet strings that begins with 2 consonants and end with 3 vowels = 5259544316

7 0

3 years ago

This system of equations has A. exactly one solution B. no solution C. infinitely many solution

german

Answer:

C.

Step-by-step explanation:

This is because both lines are parallel, leaving infinite solutions.

5 0

3 years ago

The width of a rectangular painting is 5 inches shorter than it's length. Creat a table of values to include possible widths,len

Misha Larkins [42]

A = L * W
W = L - 5

these are in inches...
length width area
10 5 50 in^2
9 4 36 "
8 3 24 "
7 2 14 "
6 1 6 "

6 0

4 years ago

CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is

Rufina [12.5K]

Answer:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

Step-by-step explanation:

We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

$y=\sin(x)\cos(x)$

Take the derivative of both sides with respect to x:

$\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]$

We need to use the product rule:

$(uv)'=u'v+uv'$

So, differentiate:

$y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]$

Evaluate:

$y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))$

Simplify:

$y'=\cos^2(x)-\sin^2(x)$

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

$0=\cos^2(x)-\sin^2(x)$

Now, let's solve for x. First, we can use the difference of two squares to obtain:

$0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))$

Zero Product Property:

$0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)$

Solve for each case.

Case 1:

$0=\cos(x)-\sin(x)$

Add sin(x) to both sides:

$\cos(x)=\sin(x)$

To solve this, we can use the unit circle.

Recall at what points cosine equals sine.

This only happens twice: at π/4 (45°) and at 5π/4 (225°).

At both of these points, both cosine and sine equals √2/2 and -√2/2.

And between the intervals 0 and 2π, these are the only two times that happens.

Case II:

We have:

$0=\cos(x)+\sin(x)$

Subtract sine from both sides:

$\cos(x)=-\sin(x)$

Again, we can use the unit circle. Recall when cosine is the opposite of sine.

Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

$\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}$

And we're done!

Edit: Small Mistake :)

5 0

3 years ago

Help me resolve this 3.25 × 6.5 =

Alexeev081 [22]

The answer is:

21.125

8 0

3 years ago

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