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4 years ago

12

4(x − 4) + 2(3x2 + 3x − 20)

1 answer:

Feliz [49]4 years ago

7 0

Answer:

$10x-56 + 6x^{2}$

Step-by-step explanation:

Step 1: $4x-16+2(3x^{2} +3x-20)$

Step 2: $4x-16+6x^{2} +6x-40$

Step 3: $(4x+6x) + (-16-40) +6x^{2}$

Step 4: $10x - 56 + 6x^{2}$

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Need help with this please

Advocard [28]

Answer:

y = 1/2x + 3

Step-by-step explanation:

when X = 0, y = 3

when y = 0, 0 = kx + 3

k = 1/2

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3 years ago

What is the area of a square figure that has a length of 5 cm?

NikAS [45]

Answer:

its 5x5x5

Step-by-step explanation:

125

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4 years ago

Read 2 more answers

Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by h(

andreev551 [17]

Answer:

a. 38.19m/s

b. 38.605m/s

c. 38.937m/s

d. 39.0117m/s

e. 39.01917m/s

Step-by-step explanation:

The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:

$v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}$

Where:

$x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval$

a. Let's find h(3) and h(4) using the data provided by the problem:

$h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f$

The average velocity over the interval [3, 4] is :

$v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s$

b. Let's find h(3.5) using the data provided by the problem:

$h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f$

The average velocity over the interval [3, 3.5] is :

$v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s$

c. Let's find h(3.1) using the data provided by the problem:

$h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f$

The average velocity over the interval [3, 3.1] is :

$v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s$

d. Let's find h(3.01) using the data provided by the problem:

$h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f$

The average velocity over the interval [3, 3.01] is :

$v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s$

e. Let's find h(3.001) using the data provided by the problem:

$h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f$

$v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s$

7 0

3 years ago

Find the limit of the sequence of partial sums whose general term is <img src="https://tex.z-dn.net/?f=a_n%3D%5Cfrac%7B100%5En%7

nasty-shy [4]

Answer:

0

Step-by-step explanation:

If ∑aₙ converges, then lim(n→∞)aₙ = 0.

Using ratio test, we can determine if the series converges:

If lim(n→∞) |aₙ₊₁ / aₙ| < 1, then ∑aₙ converges.

If lim(n→∞) |aₙ₊₁ / aₙ| > 1, then ∑aₙ diverges.

lim(n→∞) |(100ⁿ⁺¹ / (n+1)!) / (100ⁿ / n!)|

lim(n→∞) |(100ⁿ⁺¹ / (n+1)!) × (n! / 100ⁿ)|

lim(n→∞) |(100 / (n+1)|

0 < 1

The series converges. Therefore, lim(n→∞)aₙ = 0.

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3 years ago

<img src="https://tex.z-dn.net/?f=%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%

White raven [17]

Answer:

[Tex] \:/:/://///////////////////////////////////////::::::::::::::::::::::::::::::::7474838383+$83+

Answer = fracfracfracfrac tex tex????????????

8 0

3 years ago