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ivanzaharov [21]
3 years ago
6

The first term of an arithmetic sequence is −12 . The common difference of the sequence is 7. What is the sum of the first 30 te

rms of the sequence? Enter your answer in the box.
Mathematics
1 answer:
Olenka [21]3 years ago
8 0

Answer:

2685

Step-by-step explanation:

The nth term of an arithmetic sequence is:

aₙ = a₁ + d (n − 1)

where a₁ is the first term and d is the common difference.

Here, a₁ = -12 and d = 7:

aₙ = -12 + 7 (n − 1)

aₙ = -12 + 7n − 7

aₙ = -19 + 7n

The sum of the first n terms of an arithmetic sequence is:

S = (n/2) (a₁ + aₙ)

First, we find the 30th term:

a₃₀ = -19 + 7(30)

a₃₀ = 191

Now we find the sum:

S = (30/2) (-12 + 191)

S = 2685

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maria [59]

Answer:

a. At t = 5 s

a(5)=\frac{1785}{\left(6(5)+17\right)^2}=\frac{1785}{2209}\approx0.808 \frac{ft}{s^2}

b. At t = 10 s

a(10)=\frac{1785}{\left(6(10)+17\right)^2}=\frac{255}{847}\approx0.301 \frac{ft}{s^2}

c. At t = 20 s

a(20)=\frac{1785}{\left(6(20)+17\right)^2}=\frac{1785}{18769}\approx0.095 \frac{ft}{s^2}

Step-by-step explanation:

We know that the velocity function is given by

                                                   v(t)=\frac{105t}{6t+17}

Acceleration is the rate of change of velocity so we take the derivative of the velocity function with respect to time.

a(t)=\frac{dv}{dt}=\frac{d}{dt} (\frac{105t}{6t+17})

\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\105\frac{d}{dt}\left(\frac{t}{6t+17}\right)\\\\\mathrm{Apply\:the\:Quotient\:Rule}:\quad \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{\frac{d}{{dx}}f\left( x \right)g\left( x \right) - f\left( x \right)\frac{d}{{dx}}g\left( x \right)}}{{g^2 \left( x \right)}}

105\cdot \frac{\frac{d}{dt}\left(t\right)\left(6t+17\right)-\frac{d}{dt}\left(6t+17\right)t}{\left(6t+17\right)^2}\\\\105\cdot \frac{1\cdot \left(6t+17\right)-6t}{\left(6t+17\right)^2}\\\\a(t)=\frac{1785}{\left(6t+17\right)^2}

a. At t = 5 s

a(5)=\frac{1785}{\left(6(5)+17\right)^2}=\frac{1785}{2209}\approx0.808 \frac{ft}{s^2}

b. At t = 10 s

a(10)=\frac{1785}{\left(6(10)+17\right)^2}=\frac{255}{847}\approx0.301 \frac{ft}{s^2}

c. At t = 20 s

a(20)=\frac{1785}{\left(6(20)+17\right)^2}=\frac{1785}{18769}\approx0.095 \frac{ft}{s^2}

3 0
3 years ago
The equation y minus 1 = negative 7 (x minus 3). is written in point-slope form. What is the y-intercept of the line?
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Answer:

(0, 22)

Step-by-step explanation:

So you have the equation:

y-1=-7(x-3)

So first let's define what the "y-intercept" is. The y-intercept, of a line is like the name implies, when the line intercepts or "passes" the y-axis. And if you were to take any point on the y-axis, the y-value may vary, but there is one value that doesn't change, and that's the x-value. the x-value is always going to be zero, no matter where you are on the y-axis. So to find the y-intercept simple plug 0 in as x. This gives you the equation:

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y-1 = -7(-3)

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So the y-intercept is at (0, 22)

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