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3 years ago

(a^2 + 3a - 18 / a^2 - 5a +6)(5a - 10 / a^2 -36)

1 answer:

8 0

$\frac{a^{2}+3a-18}{a^2-5a+6} * \frac{5a-10}{a^{2}-36} = \frac{(a+6)(a-3)}{(a-2)(a-3)} * \frac{5(a-2)}{(a+6)(a-6)} = \frac{5}{a-6} ,

where a \neq 2, a \neq 3, a \neq +6,a \neq -6$

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The answer is 2.5 * 10^6

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3 years ago

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4 years ago

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MArishka [77]

Answer:

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Step-by-step explanation:

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3 years ago

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zaharov [31]

Answer:

3. The vertex form of the function, f(x) = x² - 4·x - 17 is f(x) = (x - 2)² - 21

4. The solutions are, x = -2 + √10 and x = -2 - √10

5. The quadratic equation with vertex (3, 1) and a = 1 in standard form is given as follows;

f(x) = x² - 6·x + 10

Step-by-step explanation:

3. The function given in standard form is f(x) = x² - 4·x - 17, which is the form, f(x) = a·x² + b·x + c

The vertex form of the of a quadratic function can be presented based on the above standard form as follows;

f(x) = a(x - h)² + k

Where;

(h, k) = The coordinate of the vertex

h = -b/2a

k = f(h)

Comparing with the given equation, we have;

f(x) = a·x² + b·x + c = x² - 4·x - 17

a = 1

b = -4

c = -17

∴ h = -(-4)/(2 × 1) = 2

h = 2

k = f(h) = f(2) = 2² - 4 × 2 - 17 = -21

k = -21

The vertex form of the function, f(x) = x² - 4·x - 17 is therefore, given as follows;

f(x) = (x - 2)² - 21

4. The given equation for which we need to solve by completing the square is 2·x² + 8·x = 12

Dividing the given equation by 2 gives;

x² + 4·x = 6

Which is of the form, x² + b·x = c

Where;

a = 1

b = 4

c = 6

From which we add (b/2)² to both sides to get x² + b·x + (b/2)² = c + (b/2)²

Adding (b/2)² = (4/2)² to both sides of x² + 4·x = 6 gives;

x² + 4·x + 4 = 6 + 4

(x + 2)² = 10

x + 2 = ±√10

x = -2 ± √10

The solution are, x = -2 + √10 and x = -2 - √10

5. Given that the value of the vertex = (3, 1), and a = 1, we have;

The vertex, (h, k) = (3, 1)

h = 3, k = 1

Therefore, h = 3 = -b/(2 × a) = -b/(2 × 1)

∴ -b = 2 × 3 = 6

b = -6

k = f(h) = a·h² + b·h + c, by substitution, we have;

k = f(3) = 1 × 3² + (-6) × 3 + c = 1

∴ c = 1 - (1 × 3² + (-6) × 3) = 10

c = 10

The quadratic equation with vertex (3, 1) and a = 1 in standard form, f(x) a·x² + b·x + c is therefor;

f(x) = x² - 6·x + 10

4 0

3 years ago

A movie projector positioned 34 feet from a wall creates an image that is 8 feet wide on the wall. if a screen is placed 5 fee

inessss [21]

Answer: 1.48ft

Step-by-step explanation:

The angle, θ, at which the projector "spreads" the image is constant.

We know that:

"A movie projector positioned 34 feet from a wall creates an image that is 8 feet wide on the wall"

We can think this problem as the image in the end, two triangle rectangles with cathetus equal to 34 ft, and 8/2 ft = 4ft

Then this angle θ can be calculated because we know the two cathetus of this triangle:

Tan(θ) = (opposite cathetus)/(adjacent cathetus)

Tan(θ) = 4ft/34ft

θ = Cotan(4/34) = 8.46°

Now, if you move the place the projector 5ft from the wall, now the adjacent cathetus is 5ft instead of 34.

Then we have:

Tan(8.46°) = x/5ft

Tan(8.46°)*5ft = x = 0.74ft

But remember that the actual width of the image will be two times that, so the width is:

W = 2*0.74ft = 1.48ft

5 0

4 years ago