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2 years ago

Simplify this radical expression square root of 4w

Mathematics

1 answer:

dexar [7]2 years ago

3 0

2w maybe? The square root of four is two...

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A new car worth $20,000 loses 20% of its value every year. Is the value of the car represented by a linear or exponential functi

madreJ [45]

Exponential.

One way to think about this, is to ask how much you will lose each year. The first year you're losing 20% of $20,000 ($4,000). The second year you're losing 20% of 16,000 ($3,200). The dollar value your car loses each year diminishes exponentially, so it is not a linear but an exponential function.

7 0

3 years ago

Which is a zero if the quadratic function f(x) = 16x2 + 32x - 9

Artist 52 [7]

x= 1/4 , -9/4

when you set the function to equal to zero you will get these values

5 0

3 years ago

Solve the system of equations graphicallySolution=?

BigorU [14]

The given system is

$\begin{cases}y=\frac{5}{2}x-8 \\ y=-\frac{1}{2}x-2\end{cases}$

Given that y = y, let's combine the equations

$\frac{5}{2}x-8=-\frac{1}{2}x-2$

Let's solve for x

$\begin{gathered} \frac{5}{2}x+\frac{1}{2}x=8-2 \\ \frac{6}{2}x=6 \\ 3x=6 \\ x=\frac{6}{3} \\ x=2 \end{gathered}$

Then, we find y

$y=-\frac{1}{2}x-2=-\frac{1}{2}\cdot2-2=-1-2=-3$

The solution is (2, -3).

The image below shows the solution graphically.

3 0

1 year ago

Sam drove 715 miles in 11 hours. At the same rate, how many miles would he drive in 9 hours?

KATRIN_1 [288]

Answer:

585 miles

Step-by-step explanation:

8 0

3 years ago

EXPLAIN why we placed the value of x= 4/3( the minimum value) into the equ of gradient(dy/dx) [in the answer, marking scheme att

aliina [53]

$y=x(x-2)^2$
$\implies y'=(x-2)^2+2x(x-2)=3x^2-8x+4=(3x-2)(x-2)=0$
$\implies x=\dfrac23,x=2$

are the critical points, and judging by the picture alone, you must have $b=\dfrac23$ and $a=2$ . (You might want to verify with the derivative test in case that's expected.)

Then the shaded region has area

$\displaystyle\int_0^2x(x-2)^2\,\mathrm dx=\dfrac43$

I'll leave the details to you.

Now, for part (iv), you're asked to find the minimum of $\dfrac{\mathrm dy}{\mathrm dx}=y'$ , which entails first finding the second derivative:

$y'=3x^2-8x+4$
$\implies y''=6x-8$

setting equal to 0 and finding the critical point:

$6x-8=0\implies x=\dfrac86=\dfrac43$

This is to say the minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ *occurs when $x=\dfrac43$ *, but this is not necessarily the same as saying that $\dfrac43$ is the actual minimum value.

The minimum value of $\dfrac{\mathrm dy}{\mathrm dx}$ is obtained by evaluating the derivative at this critical point:

$m=\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=4/3}=3\left(\dfrac43\right)^2-8\left(\dfrac43\right)+4=-\dfrac43$

4 0

3 years ago