All categories

1 year ago

Check off the methods of differentiation required, and complete the derivative.

Mathematics

1 answer:

kogti [31]1 year ago

6 0

Given:

$f(x)=\sqrt[]{5x^2-2}$

f(X) will be differentiated by chain rule.

$\begin{gathered} f^{\prime}(x)=\frac{1}{2\sqrt[]{5x^2-2}}\lbrack5(2x)\rbrack \\ f^{\prime}(x)=\frac{5x}{\sqrt[]{5x^2-2}} \end{gathered}$

You might be interested in

Asphere has a radius of 27 inches. A horizontal plane passes through the center of the sphere.

Westkost [7]

9514 1404 393

Answer:

circle of radius 27 inches

Step-by-step explanation:

Anywhere a plane cuts a sphere, the cross section is a circle. When the plane includes the center of the sphere, the circle has the same radius the sphere has.

The cross section is a circle of radius 27 inches.

7 0

3 years ago

*NEED ANSWER QUICK!!*

Rom4ik [11]

Answer:

You would need 9/4 or 2 1/4 gallons of paint

Step-by-step explanation:

You would need 2 1/4 gallon of paint because if 1/3 of the room takes 1/4 paint, then 3/4 would finish a room, multiply the three by three for three rooms and there you have 9/4 or 2 1/4 gallons

6 0

2 years ago

Susan needs to simplify the expression shown below.

mash [69]

Answer:

The answer to 75 - ( 8 + 45 ÷ 3 ) × 2= 29

3 0

3 years ago

Describe a pattern you notice when adding the sums of fractions with even denominators as opposed to those with odd denominators

GuDViN [60]

When adding fractions with even denominators, you will not get a whole number, there will be a 1/2. When adding fractions with odd denominators there will be a whole number.

8 0

3 years ago

Read 2 more answers

Find the distance from P to l.

ki77a [65]

$Equation\ of\ a\ line\ from\ 2\ points (x_1;\ y_1);\ (x_2;\ y_2):\\\\y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\(-4;\ 2)\ and\ (3;-5)\\subtitute:\\\\y-2=\dfrac{-5-2}{3-(-4)}\cdot(x-(-4))\\\\y-2=\dfrac{-7}{7}\cdot(x+4)\\\\y-2=-(x+4)\\\\y-2=-x-4\ \ \ |add\ x\ and\ 4\ to\ both\ sides\\\\x+y+2=0\\-------------------\\$

$Point-line\ distance\\l:Ax+By+C=0;\ (x_0;\ y_0)\\\\d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\\\\x+y+2=0\to A=1;\ B=1;\ C=2\\(1;\ 2)\to x_0=1;\ y_0=2\\subtitute\\\\d=\dfrac{|1\cdot1+1\cdot2+2|}{\sqrt{1^2+1^2}}=\dfrac{|1+2+2|}{\sqrt2}=\dfrac{|5|}{\sqrt2}=\dfrac{5}{\sqrt2}=\dfrac{5\cdot\sqrt2}{\sqrt2\cdot\sqrt2}\\\\=\boxed{\dfrac{5\sqrt2}{2}}$

4 0

3 years ago