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2 years ago

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Todor was trying to factor 10x^2-5x+15. He found that the greatest common factor of these terms was 5 and made an area model: Wh

at is the width of Todor's area model?

1 answer:

jenyasd209 [6]2 years ago

4 0

Answer:

The width of the area model is equal to

$(2x^2-x+3)\ units$

Step-by-step explanation:

we know that

The area of a rectangular model is given by the formula

$A=LW$ ----> equation A

where

L is the length

W is the width

we have

$A=10x^2-5x+15$

Factor the expression

$A=5(2x^2-x+3)$

substitute the value of the Area in the equation A

$5(2x^2-x+3)=LW$

In this problem

The greatest common factor of these terms is the length (L=5 units)

so

we can say that the width is equal to (2x^2-x+3)

therefore

The width of the area model is equal to

$(2x^2-x+3)\ units$

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the jellybean jar has a radius of 4.8 cm and a height of 11.8 cm. what would be a reasonable lower limit for the number of jelly

azamat

Answer:

$\large\boxed{\large\boxed{100}}$

Explanation:

This question comes with four answer choices:

10,000
100,000
100
1

You just need to find a rough estimate of the volume of the jar and the volume of one jelly bean. This is called magnitude order.

The <em>jellybean ja</em>r is cylindrical, Thus, its volume is $\pi \times radius^2\times height$

To find the order of magnitude, you just use the numbers rounded to one signficant figure: round π to 3, the radius to 5, cm, and the height to 10:

$Volume\approx 3cm\times (5cm)^2\times10cm=750cm^3\approx1,000cm^3$

The order of magnitude for the radius of a jellybean is 1 cm. And the order of magnitude of the volume of a sphere with a radius of 1 cm is the cube of the diameter (2cm):

$(2cm)^3=8cm^3\approx10cm^3$

Hence, a reasonable lower limit for the number of jellybeans in the jar is:

$1,000cm^3/10cm^3=100$

7 0

2 years ago

Pluto's distance P(t)P(t)P, left parenthesis, t, right parenthesis (in billions of kilometers) from the sun as a function of tim

Xelga [282]

Answer: P(t) = 1.25.sin( $\frac{\pi}{3}$ .t) + 5.65

Step-by-step explanation: A motion repeating itself in a fixed time period is a periodic motion and can be modeled by the functions:

y = A.sin(B.t - C) + D or y = Acos(B.t - C) + D

where:

A is amplitude A=|A|

B is related to the period by: T = $\frac{2.\pi}{B}$

C is the phase shift or horizontal shift: $\frac{C}{B}$

D is the vertical shift

In this question, the motion of Pluto is modeled by a sine function and doesn't have phase shift, C = 0.

<u>Amplitude</u>:

a = $\frac{largest - smallest}{2}$

At t=0, Pluto is the farthest from the sun, a distance 6.9 billions km away. At t=66, it is closest to the star, P(66) = 4.4 billions km. Then:

a = $\frac{6.9-4.4}{2}$

a = 1.25

<u>b</u>

A time period for Pluto is T=66 years:

66 = $\frac{2.\pi}{b}$

b = $\frac{\pi}{33}$

<u>Vertical</u> <u>Shift</u>

It can be calculated as:

d = $\frac{largest+smallest}{2}$

d = $\frac{6.9+4.4}{2}$

d = 5.65

Knowing a, b and d, substitute in the equivalent positions and find P(t).

P(t) = a.sin(b.t) + d

P(t) = 1.25.sin( $\frac{\pi}{3}$ .t) + 5.65

The Pluto's distance from the sun as a function of time is

P(t) = 1.25.sin( $\frac{\pi}{3}$ .t) + 5.65

8 0

2 years ago

Bill, George, and Ross, in order, roll a die. The first one to roll an even number wins and the game is ended. What is the proba

hram777 [196]

Answer: $0.5714\frac{1}{15}$

Step-by-step explanation:

Given that Bill, George, and Ross, in order, roll a die.

The first one to roll an even number wins and the game is ended.

Since Bill starts the game he can win by throwing even number or lose by throwing odd number

P(win) = 0.5, otherwise, the die will go to George. For Bill to win, both George and Ross should throw an odd number so that Bill again gets the chance with game non ending.

Thus we have Prob of Bill winning =P of Bill winning in I throw +P of Bill winning in his II chance of throw +....infinitely

To get back the dice once he loses probability

= p both throws odd = $0.5(0.5) = 0.25$

Thus Prob for Bill winning

= $0.5+0.25(0.5)^2+0.25(0.5)^3(0.25)+...$

This is an infinite geometric series with I term 0.5 and common ratio 0.125<1

Sum = $\frac{a}{1-r} =\frac{0.5}{1-0.125} =0.5714$

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2 years ago

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Dennis_Churaev [7]

400+200 or add 368+231= 599 and you round your answer, woulb be 600

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jenyasd209 [6]

Answer:

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Step-by-step explanation:

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2 years ago

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