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grin007 [14]
2 years ago
5

Find a number halfway between 2 1/2 and 1 1/3

Mathematics
1 answer:
kvasek [131]2 years ago
4 0

Answer:

23/12

Step-by-step explanation:

2 1/2 = 5/2(3/3) = 15/6

1 1/3 = 4/3(2/2) = 8/6

15/6 - 8/6 = 7/6

7/6(1/2) = 7/12

16/12 + 7/12 = 23/12

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Elbert made 48% of the shots he took during the basketball game. If he attempted 25 shots, how many times did he score?
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Hi there! So Elbert made 48% of his shots. To find out how many shots he made, all you have to do is multiply the amount of shots he attempted by the percentage. 25 * 48% is 12. There. Elbert scored 12 baskets.
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3 years ago
Find the product: (n-5)(n-1)
hjlf

Answer:

6n=5

Step-by-step explanation:

hope it helps tell me how you do

7 0
3 years ago
How do you get this answer ?
Leno4ka [110]
If you 
Let x to be the total number of pages in a book,

then 1/3x = 117

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3 years ago
Solve the following equation:<br> 1/3(y - 2) -5/6 (y + 1) =3/ 4(0 - 3) - 2
Evgesh-ka [11]

Answer:

11/5

Step-by-step explanation:

If you need help with more problems like this you can use m a t h w a y. (remove the spaces)  :)

8 0
3 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
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