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3 years ago

5

Daniela y Jorge can en bicicleta y salen del mismo lugar, Daniela avanza 6km y luego retrocede 2km mientras que Jorge avanza 8km

y retrocede 5km.
¿A que distancia se encuentra uno del otro?

¿Quien ha avanzado más de los dos?

¿Quien ha recorrido más km?

1 answer:

den301095 [7]3 years ago

7 0

Answer:

a) 1 km

b) Amaya

c) Jorge

HOPE THIS HELPS ;P

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(Show your working) I uploaded the question

svlad2 [7]

x^2 = the first integer

(x - 1)^2 = the second integer.

x^2 - (x - 1)^2 = ?

First, let's plug a number into our equation for x.

(2)^2 - (2 - 1)^2 = ?

4 - (1)^2 = ?

4 - 1 = 3

As we can see the difference is odd but it's also the sum of the two consecutive integers.

2 + 1 = 3.

This works for all numbers. Let's plug another number into our equation for x.

(4)^2 - (4 - 1)^2 = ?

16 - (3)^2 = ?

16 - 9 = 7

4 + 3 = 7

Try any number and it will always be odd.

4 0

3 years ago

By coincidence, Kimberly's Paint Supplies sold equal quantities of two paint colors yesterday: yellow and blue. All yellow paint

Vlad1618 [11]

Answer:

36 liters

Step-by-step explanation:

We solve this question using the least common multiple method

Find and list multiples of the liter the blue paint comes in the liters the yellow paint comes in. We find the first common multiple is found. This is the least common multiple.

Multiples of 9:

9, 18, 27, 36, 45, 54

Multiples of 12:

12, 24, 36, 48, 60

Therefore,

LCM(9, 12) = 36

Hence, the smallest amount of each paint color the store must have sold is 36 liters

8 0

3 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

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