What is 5 divided by 535 long divison

A chord of a circle is 66cm long, it is 56 cm from the center of the circle. What is the radius of the circle?

ser-zykov [4K]

Answer:

radius = 65 cm

Step-by-step explanation:

(See image attached for better comprehension)

To find the value of the radius, we just need to draw a triangle, where one side is half of the chord (the distance from the chord to the center creates a point that divides the chord in two equal segments), other side is the distance from the chord to the center, and the hypotenusa will be the radius.

So, using Pythagoras' theorem, we have that:

r^2 = 56^2 + 33^2

r^2 = 4225

r = 65 cm

4 0

3 years ago

Write the equation of a square root function that has the following graph.

Anastaziya [24]

9514 1404 393

Answer:

g(x) = √(x -5) +7

Step-by-step explanation:

To translate the graph of f(x) by h units horizontally and k units vertically, the function is transformed to ...

g(x) = f(x -h) +k

To translate f(x) = √x by 5 units horizontally and 7 units vertically, the function is transformed to ...

g(x) = √(x -5) +7

6 0

2 years ago

What theorem shows that ΔACE ≅ ΔBCD?

jolli1 [7]

Answer:

sas

Step-by-step explanation:

3 0

3 years ago

A family plans to water ski at a local lake. If the temperature outside is 30 degrees C, what is the temperature in degrees F? U

Sphinxa [80]

30 degrees C = 86 degrees F

8 0

3 years ago

Read 2 more answers

The graph of f(x)= 3/1+x^2 is shown in the figure to the right. Use the second derivative of f to find the intervals on which f

GenaCL600 [577]

Answer:

Concave Up Interval: $(- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)$

Concave Down Interval: $(\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )$

General Formulas and Concepts:

Calculus

Derivative of a Constant is 0.

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Second Derivative Test:

Possible Points of Inflection (P.P.I) - Tells us the possible x-values where the graph f(x) may change concavity. Occurs when f"(x) = 0 or undefined
Points of Inflection (P.I) - Actual x-values when the graph f(x) changes concavity
Number Line Test - Helps us determine whether a P.P.I is a P.I

Step-by-step explanation:

Step 1: Define

$f(x)=\frac{3}{1+x^2}$

Step 2: Find 2nd Derivative

1st Derivative [Quotient/Chain/Basic]: $f'(x)=\frac{0(1+x^2)-2x \cdot 3}{(1+x^2)^2}$
Simplify 1st Derivative: $f'(x)=\frac{-6x}{(1+x^2)^2}$
2nd Derivative [Quotient/Chain/Basic]: $f"(x)=\frac{-6(1+x^2)^2-2(1+x^2) \cdot 2x \cdot -6x}{((1+x^2)^2)^2}$
Simplify 2nd Derivative: $f"(x)=\frac{6(3x^2-1)}{(1+x^2)^3}$

Step 3: Find P.P.I

Set f"(x) equal to zero: $0=\frac{6(3x^2-1)}{(1+x^2)^3}$

Case 1: f" is 0

Solve Numerator: $0=6(3x^2-1)$
Divide 6: $0=3x^2-1$
Add 1: $1=3x^2$
Divide 3: $\frac{1}{3} =x^2$
Square root: $\pm \sqrt{\frac{1}{3}} =x$
Simplify: $\pm \frac{\sqrt{3}}{3} =x$
Rewrite: $x= \pm \frac{\sqrt{3}}{3}$

Case 2: f" is undefined

Solve Denominator: $0=(1+x^2)^3$
Cube root: $0=1+x^2$
Subtract 1: $-1=x^2$

We don't go into imaginary numbers when dealing with the 2nd Derivative Test, so our P.P.I is $x= \pm \frac{\sqrt{3}}{3}$ (x ≈ ±0.57735).

Step 4: Number Line Test

See Attachment.

We plug in the test points into the 2nd Derivative and see if the P.P.I is a P.I.

x = -1

Substitute: $f"(x)=\frac{6(3(-1)^2-1)}{(1+(-1)^2)^3}$
Exponents: $f"(x)=\frac{6(3(1)-1)}{(1+1)^3}$
Multiply: $f"(x)=\frac{6(3-1)}{(1+1)^3}$
Subtract/Add: $f"(x)=\frac{6(2)}{(2)^3}$
Exponents: $f"(x)=\frac{6(2)}{8}$
Multiply: $f"(x)=\frac{12}{8}$
Simplify: $f"(x)=\frac{3}{2}$

This means that the graph f(x) is concave up before $x=\frac{-\sqrt{3}}{3}$ .

x = 0

Substitute: $f"(x)=\frac{6(3(0)^2-1)}{(1+(0)^2)^3}$
Exponents: $f"(x)=\frac{6(3(0)-1)}{(1+0)^3}$
Multiply: $f"(x)=\frac{6(0-1)}{(1+0)^3}$
Subtract/Add: $f"(x)=\frac{6(-1)}{(1)^3}$
Exponents: $f"(x)=\frac{6(-1)}{1}$
Multiply: $f"(x)=\frac{-6}{1}$
Divide: $f"(x)=-6$

This means that the graph f(x) is concave down between and .

x = 1

Substitute: $f"(x)=\frac{6(3(1)^2-1)}{(1+(1)^2)^3}$
Exponents: $f"(x)=\frac{6(3(1)-1)}{(1+1)^3}$
Multiply: $f"(x)=\frac{6(3-1)}{(1+1)^3}$
Subtract/Add: $f"(x)=\frac{6(2)}{(2)^3}$
Exponents: $f"(x)=\frac{6(2)}{8}$
Multiply: $f"(x)=\frac{12}{8}$
Simplify: $f"(x)=\frac{3}{2}$

This means that the graph f(x) is concave up after $x=\frac{\sqrt{3}}{3}$ .

Step 5: Identify

Since f"(x) changes concavity from positive to negative at $x=\frac{-\sqrt{3}}{3}$ and changes from negative to positive at $x=\frac{\sqrt{3}}{3}$ , then we know that the P.P.I's $x= \pm \frac{\sqrt{3}}{3}$ are actually P.I's.

Let's find what actual point on f(x) when the concavity changes.

$x=\frac{-\sqrt{3}}{3}$

Substitute in P.I into f(x): $f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+(\frac{-\sqrt{3} }{3} )^2}$
Evaluate Exponents: $f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }$
Add: $f(\frac{-\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }$
Divide: $f(\frac{-\sqrt{3}}{3} )=\frac{9}{4}$

$x=\frac{\sqrt{3}}{3}$

Substitute in P.I into f(x): $f(\frac{\sqrt{3}}{3} )=\frac{3}{1+(\frac{\sqrt{3} }{3} )^2}$
Evaluate Exponents: $f(\frac{\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }$
Add: $f(\frac{\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }$
Divide: $f(\frac{\sqrt{3}}{3} )=\frac{9}{4}$

Step 6: Define Intervals

We know that before f(x) reaches $x=\frac{-\sqrt{3}}{3}$ , the graph is concave up. We used the 2nd Derivative Test to confirm this.

We know that after f(x) passes $x=\frac{\sqrt{3}}{3}$ , the graph is concave up. We used the 2nd Derivative Test to confirm this.

Concave Up Interval: $(- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)$

We know that after f(x) passes $x=\frac{-\sqrt{3}}{3}$ , the graph is concave up until $x=\frac{\sqrt{3}}{3}$ . We used the 2nd Derivative Test to confirm this.

Concave Down Interval: $(\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )$

6 0

2 years ago