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3 years ago

Jen and ted have cookies for snack time. Jen has 4 cookies. Ted has 1 less cookie than Jen. How many cookies does ted have

Mathematics

2 answers:

Kisachek [45]3 years ago

8 0

Ted has 3 cookies because you just take 1 away from 4 because ted has 1 less then jen

algol133 years ago

5 0

The answer for this problem is 3

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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]

8 0

3 years ago

Read 2 more answers

Find the domain and range of f(x) = -|x|

DochEvi [55]

Answers:
_______________________________________

Domain: {x| x = ℝ} .

Range: {y| y ≤ 0 } .
_______________________________________

7 0

3 years ago

Unfabricante de piezas para ensamblar bicicletas afirma que al menos el 95 por ciento de las piezas que fabrica, cumple con las

Mandarinka [93]

Answer:

Step-by-step explanation:

me hablo espanol

7 0

3 years ago

Need help asap please

PtichkaEL [24]

The answer is b

reason:
because i did that

8 0

3 years ago

Help plz , math equation !!

Ugo [173]

Answer:

We are given the table of a linear function

Since we have to find the equation of the function in the slope-intercept form,

we require the slope as well as the y-intercept

Finding the slope:

We know that slope of a line is Rise / Run

Since from the graph, we notice that the difference of 2 consecutive x-coordinates is 1

Hence, we have the x-coordinates at an interval of 1 and since the slope of a line is defined as the units ascended by the line after covering 1 unit in the x-coordinate

change in y = y $_{n}$ - y $_{n-1}$ (where n is the number of row in the table)

change in y = 31 - 23

change in y = 8 = Rise

Run = 1 (Since the interval in the table is 1)

Slope = Rise / Run

Slope = 8 / 1

Slope = 8

Finding the y-intercept:

We need the y-coordinate at x = 0

Going up from the third row, we notice that for every unit decreased in the x-coordinate , 8 units of the y-coordinate are decreased (since the slope =8)

In that case, the y-coordinate of every decreased x-coordinate will be 8 less than the last one

so we can say that the y-coordinate of x = 1 will be 8 less than 23 since x=1 is one less than x = 2

similarly, we can also say that the y-coordinate of x = 0 will be 8 less than (23-8) since x = 0 is one less than x = 1

Therefore, the y-coordinate at x =0

y = (23-8-8)

y-intercept = 7

Making the Equation:

We need to write the equation in the form:

y = ax + b (where a is the slope of the line and b is the y-intercept)

Replacing the known values, we get

y = 8x + 7

Therefore, the equation of the formula shown in the table is y = 8x + 7

5 0

3 years ago