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Dmitrij [34]
3 years ago
7

Fill in the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions: 2a

b-6a+5b+___ Please include an explanation too!
Mathematics
1 answer:
Vitek1552 [10]3 years ago
3 0

Answer:

2ab - 6a + 5b - 15

Step-by-step explanation:

Given

2ab - 6a + 5b + \_

Required

Fill in the gap to produce the product of linear expressions

2ab - 6a + 5b + \_

Split to 2

(2ab - 6a) + (5b + \_)

Factorize the first bracket

2a(b - 3) + (5b + \_)

Represent the _ with X

2a(b - 3) + (5b + X)

Factorize the second bracket

2a(b - 3) + 5(b + \frac{X}{5})

To result in a linear expression, then the following condition must be satisfied;

b - 3 = b + \frac{X}{5}

Subtract b from both sides

b - b- 3 = b - b+ \frac{X}{5}

- 3 = \frac{X}{5}

Multiply both sides by 5

- 3 * 5 = \frac{X}{5} * 5

X = -15

Substitute -15 for X in 2a(b - 3) + 5(b + \frac{X}{5})

2a(b - 3) + 5(b + \frac{-15}{5})

2a(b - 3) + 5(b - \frac{15}{5})

2a(b - 3) + 5(b - 3)

(2a + 5)(b - 3)

The two linear expressions are (2a+ 5) and (b - 3)

Their product will result in 2ab - 6a + 5b - 15

<em>Hence, the constant is -15</em>

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viva [34]

Answer:

\frac{3x}{3 - 2x}

Step-by-step explanation:

Here,

f(x) = y

y =  \frac{3x}{2x + 3}

or , swapping x with y

x =  \frac{3y}{2y + 3}

now to solve for y we get

y =  \frac{3x}{3 - 2x}

now we put f inverse x instead of y

{f}^{ - 1} (x) =  \frac{3x}{3 - 2x}

I am done.

4 0
3 years ago
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Jennifer is saving money to buy a bike. The bike costs $245. She has $125 saved, and each week she adds $15 to her savings. How
Lady bird [3.3K]
It will take her 8 weeks to have enough money for a bike. I made the equation 125+15x=245 and multiplied 15 by 8 to get exactly 245
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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 :)

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Answer: 57.5

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Since 30 divided by 4 is 7.5, our next step will read 50 + 7.5 and 50 + 7.5 simplifies to 14.

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