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Afina-wow [57]
3 years ago
9

Simplify the algebraic expression: 14xy + xy – 4yx + yx

Mathematics
1 answer:
Marina86 [1]3 years ago
6 0

Answer:

15xy-3yx


Combine like terms:

each term with similar variables are added together.

example: 14xy and xy


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Help me pls the possible answers are 50⁰ 60⁰ 100⁰ and 240⁰​
Alex Ar [27]

9514 1404 393

Answer:

  (b)  60°

Step-by-step explanation:

The sum of angles in a pentagon is 540°, so you have ...

  90° +90° +150° +x +150° = 540°

  x = 540° -480°

  x = 60°

3 0
3 years ago
PLSSS HELP IF YOU TURLY KNOW THISS
balandron [24]

Answer:

4.117 *little10 to the -29th power

Step-by-step explanation:

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2 years ago
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32ax + 12bx - 48ay - 18by<br><br>factor the polynomial​
SashulF [63]

Answer:

(16a + 6b) (2x - 3y)      

Step-by-step explanation:

32ax + 12bx - 48ay - 18by

(32ax - 48ay) + (12bx - 18by)

16a(2x - 3y) + 6b(2x - 3y)         Factor out in both seperate expressions

(16a + 6b) (2x - 3y)                    Double factoring

4 0
3 years ago
Read 2 more answers
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
Someone please help!! I’ll give you Brainliest!
Valentin [98]

Answer:

x= 1

y= 2

(1,2)

x=2

y=3

(2,3)

x=3

y=4

(3,4)

5 0
3 years ago
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