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3 years ago

13

Can someone help me solve this? I keep getting stuck

1 answer:

astra-53 [7]3 years ago

7 0

I can't imagine why the variable in the exponent isn't 'n'.
To me, that's the only way this makes sense, and I'm
going to handle it as if the exponent is (n - 1) .

First term: n = 1

A(1) = -2 · 5¹⁻¹ = -2 · 5⁰ = -2 · 1 = -2 .

Fourth term: n = 4

A(4) = -2 · 5⁴⁻¹ = -2 · 5³ = -2 · 125 = -250 .

Eighth term: n = 8

A(8) = -2 · 5⁸⁻¹ = -2 · 5⁷ = -2 · 78,125 = -156,250 .

The first choice on the list looks like the correct one.

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A particle moves along the curve below. y = √ 8 + x 3 As it reaches the point ( 1 , 3 ) , the y -coordinate is increasing at a r

Dmitry [639]

Answer:

<em>The x-coordinate is changing at 10 cm/s</em>

Step-by-step explanation:

<u>Rate of Change</u>

Suppose two variables x and y are related by a given function y=f(x). If they both change with respect to a third variable (time, for instance), the rate of change of them is computed as the derivative using the chain rule:

$\displaystyle \frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}$

We have

$y=\sqrt{8+x^3}$

Or, equivalently

$y^2=8+x^3$

We need to know the rate of change of x respect to t. We'll use implicit differentiation:

$\displaystyle 2y\cdot \frac{dy}{dt}=3x^2\cdot \frac{dx}{dt}$

Solving for dx/dt

$\displaystyle \frac{dx}{dt}=\frac{2y\cdot \frac{dy}{dt}}{3x^2}$

Plugging in the values x=1, y=3, dy/dt=5

$\displaystyle \frac{dx}{dt}=\frac{2(3)\cdot 5}{3\cdot 1^2}=10$

The x-coordinate is changing at 10 cm/s

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3 years ago

QUUESTION Today, everything at a store is on sale! The store offers a 20% discount.

julsineya [31]

Answer:

(x÷100)20

Step-by-step explanation:

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3 years ago

How do you find mode in data

nikdorinn [45]

Answer:

The mode is the number that shows up most frequently

Step-by-step explanation:

Example, in the number set below

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4 0

3 years ago

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The graph of $y=ax^2+bx+c$ passes through points $(0,5)$, $(1,10)$, and $(2,19)$. Find $a+b+c$.

MAVERICK [17]

Answer:

$a+b+c=10$

Step-by-step explanation:

We are given that the graph of the equation:

$y=ax^2+bx+c$

Passes through the three points (0, 5), (1, 10), and (2, 19).

And we want to find the value of (<em>a</em> + <em>b</em> + <em>c</em>).

First, since the graph passes through (0, 5), its <em>y-</em>intercept or <em>c</em> is 5. Hence:

$y=ax^2+bx+5$

Next, since the graph passes through (1, 10), when <em>x</em> = 1, <em>y</em> = 10. Substitute:

$(10)=a(1)^2+b(1)+5$

Simplify:

$5=a+b$

The point (2, 19) tells us that when <em>x</em> = 2, <em>y</em> = 19. Substitute:

$(19)=a(2)^2+b(2)+5$

Simplify:

$14=4a+2b$

This yields a system of equations:

$\begin{cases} 5 = a + b \\ 14 = 4a + 2b\end{cases}$

Solve the system. We can do so using elimination (or any other method you prefer). Multiply the first equation by negative two:

$-10=-2a-2b$

Add the two equations together:

$(-10)+(14)=(-2a+4a)+(-2b+2b)$

Combine like terms:

$4 = 2a$

Hence:

$a=2$

Using the first equation:

$5=(2)+b\Rightarrow b=3$

Therefore, our equation is:

$y=2x^2+3x+5$

Thus, the value of (<em>a</em> + <em>b</em> + <em>c</em>) will be:

$a+b+c = (2) + (3) + (5) = 10$

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Lena [83]

Answer:

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Step-by-step explanation:

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