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kap26 [50]
3 years ago
10

HELP ASAP one solution to a quadratic function, f, is given below. (sqrt of 7) + 5i which of the following statements is true ab

out the given function? a) the other solution to the function is - sqrt7 -5i b) the other solution to the function is sqrt7 -5i c) the other solution is -sqrt7 + 5i d) function f has no other solutions.
Mathematics
2 answers:
mash [69]3 years ago
7 0

Answer:

b

Step-by-step explanation:

complex solutions occur in conjugate pairs, hence

\sqrt{7} + 5i , then \sqrt{7} - 5i is also a solution


mel-nik [20]3 years ago
4 0

Answer:

Option b). the other solution to the function is \sqrt{7}-5i

Step-by-step explanation:

In a given quadratic function f(x) = ax² + bx + c, if b² - 4ac < 0 then bothe the solutions of the function will have two complex roots.

Roots of this function will be the complex conjugates.

Since one root of the equation is given as \sqrt{7}+5i

Then other solution will be its complex conjugate as \sqrt{7}-5i

Therefore, Option B) will be the answer.

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Two collinear points on a line are given in the table below:
katrin [286]

Answer:

(4,3) and (7,2) do not lie on the line

Step-by-step explanation:

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(0,0)\ and\ (2,1)

Required

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y - y_1 = m(x -x_1)

Where

m = \frac{1}{2}

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y - y_1 = m(x -x_1) becomes

y - 0 = \frac{1}{2}(x - 0)

y = \frac{1}{2}x

To determine which point is on the line, we simply plug in the  values of x to in the equation check.

For (4,2)

x = 4 and y =2

Substitute 4 for x and 2 for y in y = \frac{1}{2}x

2 = \frac{1}{2} * 4

2 = \frac{4}{2}

2=2

<em>This point is on the graph</em>

<em></em>

For (4,3)

x = 4 and y = 3

Substitute 4 for x and 3 for y in y = \frac{1}{2}x

3 = \frac{1}{2} * 4

3 = \frac{4}{2}

3 \neq 2

<em>This point is not on the graph</em>

<em></em>

For (7,2)

x = 7 and y = 2

Substitute 7 for x and 2 for y in y = \frac{1}{2}x

2 = \frac{1}{2} * 7

2 = \frac{7}{2}

2 \neq 3.5

<em></em>

<em>This point is not on the graph</em>

<em></em>

<em></em>(\frac{4}{8},\frac{2}{8})<em></em>

<em></em>x = \frac{4}{8} and<em> </em>y = \frac{2}{8}<em></em>

<em>Substitute </em>\frac{4}{8}<em> for x and </em>\frac{2}{8}<em> for y in </em>y = \frac{1}{2}x<em></em>

<em></em>\frac{2}{8} = \frac{1}{2} * \frac{4}{8}<em></em>

<em></em>\frac{2}{8} = \frac{1 * 4}{8 * 2}<em></em>

<em></em>\frac{2}{8} = \frac{4}{16}<em></em>

<em></em>\frac{1}{4} = \frac{1}{4}<em></em>

<em></em>

<em>This point is on the graph</em>

3 0
3 years ago
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