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3 years ago

What is the domain and range of the function f(x) = x^2 -4x-5?

1 answer:

5 0

F(x) = x²-4x-5, quadratic function,
Domain (the values if x) is all real numbers.

To find range we should draw a graph or to write an equation in vertex form.
f(x) = x²-4x+4-4-5
f(x) = (x-2)²-9
Point (-2,-9) is the vertex of the parabola, and it is a minimum because a parabola has positive sign in front of x², so it is looking up. Minimum value of y =-9
Range(the values of y) is [-9, ∞)

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How to work out:<br>5x+1=2x-8

Helen [10]

$5x+1=2x-8 \\ 5x-2x=-8+1 \\ 3x=-7 \\ x=-7/3$

<u>S= {-7/3}</u>

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3 years ago

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OAB is a sector of a circle <br>what is the area of the sector

oee [108]

Answer:

12.56 square cm

Step-by-step explanation:

$Central \:angle \:(\theta) = 40°\\Radius \: (r) = 6 cm$

Let A be the area of the sector of circle.

$\therefore \: A = \frac{\theta }{360 \degree} \times \pi {r}^{2} \\ \\= \frac{ 40 \degree}{360 \degree} \times 3.14\times {6}^{2} \\ \\ = \frac{ 1}{9} \times 3.14 \times 36 \\ \\ = 3.14 \times 4 \\ \\ = 12.56 \: {cm}^{2}$

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3 years ago

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Help

Flauer [41]

A) Measure of Angle L is 67.5. Angle F & Angle L are alternate interior angles.

b) Measure of Angle E is 112.5, because we know that F and H are vertical angles, so therefore they’re congruent. We also know H and E are adjacent, so H + E = 180.

c) Angles E and K are alternate exterior angles. Measure of Angle K is 112.5.

d) Measure of Angle H is 67.5, because we know that Angle F, which is vertical to Angle H, is 67.5. Vertical angles are congruent.

e) Measure of Angle I is 112.5, because Angle I and Angle K are vertical angles.

f) They’re corresponding angles.

3 0

2 years ago

un propietario pide 8000 para pagar un proyecto de remodelación de una cocina los términos del préstamo son de 9.2% de interés a

laila [671]

Answer:

El propietario pagará una tasa de interés de 1472 € (736x2 años)

Step-by-step explanation:

8000 / 100 x 9.2 = 736€

736 x 2 = 1472 :)

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2 years ago

Which of the following functions f : {0, 1, 2, 3} ! {0, 1, . . . , 7} are one-to-one?

allsm [11]

Answer:

1. No.

$f(0)=0\\1^2=1; \text{then } f(1)=1\\2^2=4\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^2=9\equiv 1 \text{mod 8 }; \text{then } f(3)=1$

Since f(1)=f(3) and $1\neq 3$ then f isn't one-to-one.

2. No

$f(0)=0\\1^3=1\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3=8\equiv 0\text{ mod 8}; \text{then } f(2)=0\\3^3=27\equiv 3 \text{ mod 8}; \text{then } f(3)=3$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

3. No

$0^3-8=-8\equiv 0\text{ mod 8}; \text{then } f(0)=0\\1^3-8=-7\equiv 1\text{ mod 8}; \text{then } f(1)=1\\2^3-8=0\equiv 0 \text{ mod 8}; \text{then } f(2)=0\\3^3-8=27-8=19\equiv 3 \text{ mod 8}; \text{then } f(3)=3\\$

Since f(0)=f(2) and $0\neq 2$ then f isn't one-to-one.

4. Yes

$0^3+2*0=0; \text{then } f(0)=0\\1^3+2*1=3\equiv 3\text{ mod 8}; \text{then } f(1)=3\\2^3+2*2=8+4=12\equiv 4 \text{ mod 8}; \text{then } f(2)=4\\3^3+2*3=27+6=33\equiv 1\text{ mod 8}; \text{then } f(3)=1$

Since $f(0)\neq f(1)\neq f(2) \neq f(3)$ , then f is one-to-one

5. Since f(1)=f(3) and $1\neq 3$ then, f isn't one-to-one

3 0

3 years ago