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3 years ago

Find the 22nd term of the arithmetic sequence whose common difference is d =3 and whose first term is a_1=-20

1 answer:

4 0

1,4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49,52,55,58,61,64,67,70,73,76,79,82,85,88,91,94,97,100

f(n) = 1 + 3(n-1) = 1 + 3n - 3 = 3n -2

f(22) = 3(22)-2 = 66-2 = 64

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The height in meters of a toy rocket at t seconds can be found by the

OlgaM077 [116]

Answer: 81.2

Step-by-step explanation:

h(t) = -5t² + 40t + 1.2

h(4) = -5(4)² + 40(4) + 1.2 We plug 4 seconds as t, and solve h(4)

h(4) = -80 + 160 + 1.2

h(4) = 80 + 1.2

h(4) = 81.2

4 0

3 years ago

A summer camp is organizing a hike and needs to buy granola bars for the campers. The granola bars come in small boxes and large

Irina18 [472]

Answer:

The Answer is: There are 8 small boxes and 9 large boxes. See explanation below for variables and variable definitions.

Step-by-step explanation:

Let s = small boxes. Let b = large boxes.

s + b = 17

You can solve for s:

s = 17 - b

You can solve for b:

b = 17 - s

10 times the number of small boxes plus 24 times the number of large boxes is equal to 296 granola bars.

10s + 24b = 296

Substitute:

10(17 - b) + 24b = 296

170 - 10b + 24b = 296

14b = 296 - 170

14b = 126

b = 126 / 14 = 9 large boxes

Find the number of small boxes, s:

s = 17 - b = 17 - 9 = 8 small boxes

There are 8 small boxes and 9 large boxes.

Proof:

10(8) + 24(9) = 296

80 + 216 = 296

296 = 296

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3 years ago

Graph y<1−3x. .......................... . .

Mkey [24]

Answer:

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Step-by-step explanation:

yo digo q C es la megor opceon

4 0

3 years ago

I need help solving this problem. <img src="https://tex.z-dn.net/?f=4x%20cos%20%5E%7B-1%7D%20%282x%2B4%29-%20%5Csqrt%7B3-3%20x%5

Gnesinka [82]

Given:
$f(x)=4xcos^{-1}(2x+4)-\sqrt{3-3x^2}$

Using
$\frac{d}{dx}cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}$
we derive
$\frac{d}{dx}4xcos^{-1}(2x+4)$
$=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}$

Similarly, using
$\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$
we derive
$\frac{d}{dx}(-\sqrt{3-3x^2})$
$=\frac{3x}{\sqrt{3-3x^2}}$

Therefore, the derivative is
$f'(x)=\frac{d}{dx}(4xcos^{-1}(2x+4)-\sqrt{3-3x^2})$
$=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}+\frac{3x}{\sqrt{3-3x^2}}$

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3 years ago

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Vadim26 [7]

Answer:

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Step-by-step explanation:

graph the following lines with negative slopes and compare:

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2 years ago

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