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3 years ago

Evaluate a1 (r)^n-1 for a1 = 2, r= 3, and n = 4. A. 216 B. 18 C. 32 D. 54

Mathematics

1 answer:

DaniilM [7]3 years ago

4 0

Answer:

$54$

Step-by-step explanation:

$a_1 {r}^{n - 1} \\ 2 \times {3}^{4 - 1} \\ 2 \times {3}^{3} \\ 2 \times 3 \times 3 \times 3 \\ 2 \times 9 \times 3 \\ 18 \times 3 \\ = 54$

Hope this helps you.

Let me know if you have any other questions :-):-)

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Let f(x) = xe^-x+ ce^-x, where c is a positive constant. For what positive value of c does f have an absolute

Illusion [34]

Answer:

$c=6$

Step-by-step explanation:

The absolute maximum of a continuous function $f(x)$ is where $f'(x)=0$ . Therefore, we must differentiate the function and then set $x=-5$ and $f'(x)=0$ to determine the value of $c$ :

$f(x)=xe^{-x}+ce^{-x}$

$f'(x)=-xe^{-x}+e^{-x}-ce^{-x}$

$0=-(-5)e^{-(-5)}+e^{-(-5)}-ce^{-(-5)}$

$0=5e^{5}+e^{5}-ce^{5}$

$0=e^5(5+1-c)$

$0=6-c$

$c=6$

Therefore, when $c=6$ , the absolute maximum of the function is $x=-5$ .

I've attached a graph to help you visually see this.

7 0

3 years ago

What is a fraction of 42 is 7

kramer

$\frac{7}{42}\ of\ 42\ is\ equal\ 7\\\\check:\frac{7}{42}\cdot42=\frac{7}{42}\cdot\frac{42}{1}=\frac{7}{1}=7$

3 0

3 years ago

If you know the volume of a cylinder and that a cone has the same base and height of a cone, how would you determine the volume

ratelena [41]

Answer:

Multiply the cone's volume by 3

Step-by-step explanation:

Cylinder's Volume is

V = pi r ^2(h)

Cone's Volume is

V = 1/3 pi r^2 (h)

5 0

3 years ago

Read 2 more answers

What is the slope of the line x = −5? a) undefined b) 0 c) -5 d) 5

nlexa [21]

Learn this...it helps
HOY -- horizontal line, 0 slope, represented by y = a number
VUX -- vertical line, undefined slope, represented by x = a number

now u have : x = -5.....this is a vertical line with an undefined slope

7 0

4 years ago

Vector u has a magnitude of 7 units and a direction angle of 330°. Vector v has magnitude of 8 units and a direction angle of 30

Dafna11 [192]

Keeping in mind that x = rcos(θ) and y = rsin(θ).

we know the magnitude "r" of U and V, as well as their angle θ, so let's get them in standard position form.

$\bf u=
\begin{cases}
x=7cos(330^o)\\
\qquad 7\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{7\sqrt{3}}{2}\\
y=7sin(330^o)\\
\qquad 7\cdot -\frac{1}{2}\\
\qquad -\frac{7}{2}
\end{cases}\qquad \qquad v=
\begin{cases}
x=8cos(30^o)\\
\qquad 8\cdot \frac{\sqrt{3}}{2}\\
\qquad \frac{8\sqrt{3}}{2}\\
y=8sin(30^o)\\
\qquad 8\cdot \frac{1}{2}\\
\qquad 4
\end{cases}$

$\bf u+v\implies \left( \frac{7\sqrt{3}}{2},-\frac{7}{2} \right)+\left( \frac{8\sqrt{3}}{2},4 \right)\implies \left( \frac{7\sqrt{3}}{2}+\frac{8\sqrt{3}}{2}~~,~~ -\frac{7}{2}+4\right)
\\\\\\
\left(\stackrel{a}{\frac{15\sqrt{3}}{2}}~~,~~ \stackrel{b}{\frac{1}{2}}\right)\\\\
-------------------------------$

$\bf tan(\theta )=\cfrac{b}{a}\implies tan(\theta )=\cfrac{\frac{1}{2}}{\frac{15\sqrt{3}}{2}}\implies tan(\theta )=\cfrac{1}{15\sqrt{3}}
\\\\\\
\measuredangle \theta =tan^{-1}\left( \cfrac{1}{15\sqrt{3}} \right)\implies \measuredangle \theta \approx 2.20422750397203^o$

8 0

3 years ago