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mihalych1998 [28]

3 years ago

5

PLZZZZZZZZ HELPPPPPPPPPPPP

2 answers:

polet [3.4K]3 years ago

7 0

Is that big ideas……………………..

Sindrei [870]3 years ago

5 0

Answer:

INTO THE THICK OF IT ;p

Step-by-step explanation:

You might be interested in

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago

A shopper bought a 12-pound bag of oranges for 18.75. What is the unit price per ounce?

noname [10]

Answer:

About $0.10 per ounce

Step-by-step explanation:

12 pound costs $18.75. Lets find the cost per pound first:

Cost Per Pound = $\frac{18.75}{12}=1.5625$

We know, there are 16 ounces in 1 pound.

We know 1 pound costs 1.5625

To find cost per ounce, we have to divide this by 16.

So,

Cost Per Ounce = $\frac{1.5625}{16}=0.097$

Rounded to nearest cent, that would be 10 cents per ounce

5 0

3 years ago

Solve the system algebraically.<br> 3x - 2y - 1 = 0 y = 5x + 4 <br> What is the solution?

Bogdan [553]

Answer:

$x=\frac{-9}{7}$

$y=\frac{-17}{7}$

Step-by-step explanation:

We have been given the equation:

$3x-2y-1=0$

And $y=5x+4$

Second equation can be rewritten as:

$5x-y=-4$

We will solve the equations:

$3x-2y=1$ (1)

$5x-y=-4$ (2)

Multiply equation (1) by 5 and equation (2) by 3 we get:

$15x-10y=5$ (3)

$15x-3y=-12$ (4)

Now, subtract (4) from (3) we get:

$-7y=17$

$y=\frac{-17}{7}$

Now substitute $y=\frac{-17}{7}$ in equation (1) we get:

$3x-2{\frac{-17}{7}}=1$

$3x+\frac{34}{7}=1$

$21x=-27$

$x=\frac[-27}{21}$

$x=\frac{-9}{7}$

3 0

2 years ago

Read 2 more answers

Anna hiked 95⁄8 miles in 23⁄4 hours. How fast did she walk overall? A. 31⁄4 miles per hour B. 35⁄8 miles per hour C. 31⁄2 miles

Pani-rosa [81]

Answer:

C. 3 1/2 miles per hour

Step-by-step explanation:

I just converted the fractions to decimal form.

5/8 = 0.625 9 + 0.625 = 9.625

3/4 = 0.75 2 + 0.75 = 2.75

Now divide the 2 numbers:

9.625 / 2.75 = 3.5

Finally, convert the decimal back to a fraction:

3.5 = 3 1/2

Hope this helps!

6 0

2 years ago

Read 2 more answers

A culture of bacteria has an initial population of 650 bacteria and doubles every 9 hours. Using the formula Pt=P0⋅2tdP_t = P_0\

Ghella [55]

So is doubling, first off, meaning, if the current amount say P, then when it doubles is 2P, or double that, or P + P, so the rate of growth is 100%, since it's doubling.

and is doing it every 9 hour cycle, thus

$\bf \textit{Periodic/Cyclical Exponential Growth}\\\\
A=P(1 + r)^{\frac{t}{c}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &650\\
r=rate\to 100\%\to \frac{100}{100}\to &1.00\\
t=\textit{elapsed time}\to &10\\
c=period\to &9
\end{cases}
\\\\\\
A=650(1 + 1)^{\frac{10}{9}}\implies A=650(2)^{\frac{10}{9}}\implies A=650\sqrt[9]{2^{10}}$

4 0

2 years ago