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Mamont248 [21]
3 years ago
12

MATH PLEASE HELP 50 POINTS 1 QUESTION

Mathematics
1 answer:
miv72 [106K]3 years ago
7 0
For 1, <span>∠1 and ∠5 are supplementary is a given, so you can just write 'Given'
For 2, due to that 1 and 2 are vertical angles (angles that are opposite to each other when two lines cross according to Math is Fun)
For 3, since </span>∠1=∠2, ∠1 can be substituted by ∠2 in ∠1+∠5=180 to get ∠2+∠5=180. Therefore, they are supplementary.<span>

For 4, due to that we have </span>Consecutive Interior Angles (they are angles that are on one side of a transversal, which is the line that cuts through a and b,but inside the two lines according to Math is Fun) adding up to 180 degrees, a  || b

Feel free to ask if you have any questions!
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HELPP I NEED TO TURN THIS IN 5 Mins!!!!!
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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

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