Need help with this paper!!

Which is the graph of the function f(x) = x³ + x² + x + 1?

lisabon 2012 [21]

Theres no graphs but this is how the function looks like

8 0

1 year ago

For brainily please help

OLga [1]

Answer:

The value of Z should be 11.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Please help me with this !:(

ahrayia [7]

Answer:

75.062c $m^{2}$

Step-by-step explanation:

To find the missing side use pythagorean theorem

So $6^{2} +b^2=8.5^2$

Solve for b, b= 6.021

To find the area use $\frac{1}{2}$ bh

So $\frac{1}{2}$ (6.021+19)(6)= area

And you get 75.062

7 0

2 years ago

Can someone please help me with this problem asap? Thanks!! :))

ivolga24 [154]

Answer: $\$4.34$

Step-by-step explanation:

We have the following information:

Fredburguer $F$ , milkshake $M$ and an order of fries $f$ costs $\$3.72$ :

$F+M+f=\$3.72$ (1)

One Fredburguer is equal to a milkshake plus orders of fries:

$F=M+2f$ (2)

Three milshakes is equal to a Fredburguer plus an order of fries:

$3M=F+f$ (3)

And we are asked: If we have a Fredburguer, a milkshake and two orders of fries, how much do we have to pay?

$F+M+2f=x$ (4)

Well with the first three equations we have a system of equations to solve. Let's begin by isolating $F$ from (3):

$F=3M-f$ (5)

Making (2)=(5):

$M+2f=3M-f$ (6)

Isolating $f$ :

$f=\frac{2}{3}M$ (7)

Substituting (7) in (2):

$F=M+2(\frac{2}{3}M)$ (8)

Isolating $F$ :

$F=\frac{7}{3}M$ (9)

Substituting (7) and (9) in (1):

$\frac{7}{3}M+M+\frac{2}{3}M=\$3.72$ (10)

Isolating $M$ :

$M=\$0.93$ (11) This is the cost of a Milkshake

Substituting (11) in (7):

$f=\frac{2}{3}\$0.93$ (12)

$f=\$0.62$ (13) This is the cost of an order of frieds

Substituting (11) in (9):

$F=\frac{7}{3}\$0.93$ (14)

$F=\$2.17$ (15) This is the cost of a Fredburguer

Substituting (11), (13) and (15) in (4):

$\$2.17+\$0.93+2(\$0.62)=x$ (16)

Finally:

$x=\$4.34$ This is the cost of a Fredburguer, a milkshake and two orders of fries

8 0

2 years ago

The class sizes of the introductory psychology courses at a college are shown below 121,134,106,93,149,130,119,128 the college a

frozen [14]

Center : Mean Before the introduction of the new course, center = average(121,134,106,93,149,130,119,128) = 122.5 After the introduction of the new course, center = average(121,134,106,93,149,130,119,128,45) = 113.9 The center has moved to the left (if plotted in a graph) because of the low intake for the new course. Spread before introduction of the new course : Arrange the numbers in ascending order: (93, 106,119, 121), (128, 130,134, 149) Q1=median(93,106,119,121) = 112.5 Q3=median(128,130,134,149) = 132 Spread = Interquartile range = Q3-Q1 = 19.5 After addition of the new course,

(45,93, 106,119,) 121, (128, 130,134, 149)
Q1=median(45,93,106,119)=99.5
Q3=median (128, 130,134, 149)= 132
Spread = Interquartile range = 132-99.5 =32.5
We see that the spread has increased after the addition of the new course.

3 0

3 years ago

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