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myrzilka [38]
3 years ago
7

5. The shape of a dome can be modeled by the equation h = - 2 d 2 + 100 where h is the height (in feet) of the dome from the flo

or d feet from its center. How far from the center of the dome is the height 50 feet?
6. A cab charges $0.10 per mile and a flat fee of $3.00. Write an equation to model the price y of an x-mile-long cab ride.

7. Use the discriminate to determine the number of solutions to the quadratic equation y = 3 x 2 - 4 x + 6. 8. If f ( x ) = 7 x 2 + 5, compute f ( )3 . 9. Write an equation representing the translation of f ( x ) = 7 x + 3 down 4 units.
Mathematics
1 answer:
klasskru [66]3 years ago
4 0

Answer:

5. When the dome is 50 feet high, the distance from the center is 5 feet

6. The equation that model the price y of an x-mile long ride is given as follows;

y = 0.1×x + $3.00

7. The discriminant is < 0; The equation has no real root

8. f(3) = 68

9. The equation representing the translation of the line f(x) = 7·x + 3 down 4 units is f(x) = 7·x + 7

Step-by-step explanation:

5. The given equation for the shape of the dome is presented as follows;

h = -2·d² + 100

Where;

h = The height of the dome (in feet)

d = The distance from the center

Therefore, we have;

When h = 50, d is found as follows;

h = -2·d² + 100

50 = -2·d² + 100

50 - 100 = -2·d²

-50 = -2·d²

∴ 2·d² = 50

d² = 50/2 = 25

d = √25 = 5 feet

Therefore when the dome is 50 feet high, the distance from the center is 5 feet

6. The given rate the cab charges per mile, x = $0.10

The rate the cab charges as flat fee = $3.00

Therefore, the price, y a person traveling by cab for x miles is given by the straight lie equation as y = m·x + c,

Where;

m = the slope or rate which in this case = $0.1/hour

c = A constant term which in this case = $3.00

Therefore

y = 0.1×x + $3.00

The equation that model the price y of an x-mile long ride is y = 0.1×x + $3.00

7. The discriminant, b² - 4·a·c  of the quadratic equation is (-4)² - 4×3×6 = -56

which is < 0, the equation has no real root

8. Given f(x) = 7·x² + 5

f(3) = 7 × (3)² + 5 = 68

f(3) = 68

9. The equation representing the translation of the line f(x) = 7·x + 3 down 4 units is given as follows;

Down 4 units is equivalent to subtracting 4 from the y-coordinate value, therefore, we have;

f(x) - 4= 7·x + 3

f(x) = 7·x + 3 + 4

f(x) = 7·x + 7

The equation representing the translation of the line f(x) = 7·x + 3 down 4 units is f(x) = 7·x + 7

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