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4 years ago

One solution to a quadratic function, f, is given below.

Mathematics

2 answers:

MatroZZZ [7]4 years ago

8 0

Answer: they are correct...

Step-by-step explanation: I just took the test...

Sveta_85 [38]4 years ago

7 0

The correct option is (2)

For any quadratic equation: ax2 + bx + c = 0.

The complex roots are expressed in the form a ± bi.

where a is the real part
& b is the imaginary part.
Since, one root is √7 + 5i, therefore the other root will be of the form √7 - 5i.

Hence, option (2) is correct.

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Next consider what happens when two of these functions are composed. Suppose g identifies ng regions of (0, 1)d onto (0, 1)d and

irina [24]

$n^{2}fg$ is regions of f ◦ g(·).

Step-by-step explanation:

When you multiply two functions together, you'll get a third function as the result, and that third function will be the product of the two original functions.

For example, if you multiply f(x) and g(x), their product will be h(x)=f.g(x), or h(x)=f(x)g(x).

Here we have two functions, f identifies n f regions of (0, 1)d onto (0, 1)d which is equivalent to f(x) = n f. And, g identifies n g regions of (0, 1)d onto (0, 1)d which is equivalent to g(x)= n g. Now,

⇒ ( f × g ) (x ) = f(x) × g(x)

⇒ $( fg )(x) = f(x).g(x)\\( fg )(x) = nf.ng\\(fg)(x) = n^{2}fg$

Therefore, $n^{2}fg$ is regions of f ◦ g(·).

8 0

4 years ago

N - 8 > 5 A. n > 13 B. n > 3 C. n > - 3 D. n > -13

statuscvo [17]

the answe is a- n>13

4 0

4 years ago

Match each set of points with the quadratic function whose graph passes through those points.

AnnZ [28]

Answer:

f(x) = x² - 2x - 15 passes through (-3 , 0) , (0 , -15) , (5 , 0)

f(x) = -x² - 2x + 15 passes through (-2 , 15) , (-1 , 16) , (0 , 15)

f(x) = -x² + 2x - 15 passes through (0 , -15) , (1 , -14) , (2 , -15)

Step-by-step explanation:

* Lets explain how to solve this question

- To find the points whose graph passes through them substitute the

x-coordinate in the function if the answer is the same with the

y-coordinate of the point then the graph passes through this point

lets do that

- Check the first set of points with the first function

# Pint (0 , -15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (0)² - 2(0) - 15 = -15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , -15)

# Pint (1 , -14)

∵ f(x) = x² - 2x - 15

∴ f(0) = (1)² - 2(1) - 15 = -16 ⇒ not same value of y-coordinate

∴ The graph of the function does not pass through point (1 , -14)

∴ The graph does not pass through this set of points

- Check the second set of points with the first function

# Pint (-2 , 15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (-2)² - 2(-2) - 15 = 4 + 4 - 15 -7 ⇒ not same value of y-coordinate

∴ The graph of the function does not pass through point (-2 , 15)

∴ The graph does not pass through this set of points

- Check the third set of points with the first function

# Pint (-3 , 0)

∵ f(x) = x² + 2x - 15

∴ f(0) = (-3)² - 2(-3) - 15 = 9 + 6 -15 = 0 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-3 , 0)

# Pint (0 , -15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (0)² - 2(0) - 15 = -15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , -15)

# Pint (5 , 0)

∵ f(x) = x² + 2x - 15

∴ f(0) = (5)² - 2(5) - 15 = 25 - 10 -15 = 0 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (5 , 0)

∴ The graph passes through this set of points

* f(x) = x² - 2x - 15 passes through (-3 , 0) , (0 , -15) , (5 , 0)

- Check the first set of points with the second function

# Pint (0 , -15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(0)² - 2(0) + 15 = 15 ⇒ not same value of y-coordinate

∴ The graph of the function does not passes through point (0 , -15)

∴ The graph does not pass through this set of points

- Check the second set of points with the second function

# Pint (-2 , 15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(-2)² - 2(-2) + 15 = -4 + 4 + 15 = 15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-2 , 15)

# Pint (-1 , 16)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(-1)² - 2(-1) + 15 = -1 + 2 + 15 = 16 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-1 , 16)

# Pint (0 , 15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(0)² - 2(0) + 15 = 15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , 15)

∴ The graph passes through this set of points

* f(x) = -x² - 2x + 15 passes through (-2 , 15) , (-1 , 16) , (0 , 15)

- Now we have the first set of points and the third function

∴ The graph passes through this set of points

∴ f(x) = -x² + 2x - 15 passes through (0 , -15) , (1 , -14) , (2 , -15)

7 0

4 years ago

In a city school of 900 students, 38% of the students are on the honor roll

Vera_Pavlovna [14]

Answer:

if the question is what number of 38 percent, then the answer is 342

7 0

3 years ago

Are the expressions 3x + 8 and 2x + 14 equivalent expressions for x = 6? How do you know?

siniylev [52]

Answer:

Equivalent

Step-by-step explanation:

both give 26 as answer

3 0

3 years ago