Can someone pls help me with this one problem

Round 1627187 to the nearest ten?

Eduardwww [97]

Answer:

1627190

Step-by-step explanation:

(see attached for reference)

Given the number 1627187, we can see that the number in the tens place is the number 8.

How we round this depends on the number immediately to the right of this number. (i.e the digit in the ones place)

Case 1: If the digit in the ones place is less less than 5, then the number in the tens place remains the same and replace all the digits to its right with zeros

Case 2: If the digit in the ones places is 5 or greater, then we increase the digit in the tens place and replace all the digits to its right with zeros.

In our case, the digit in the ones places is 7, this greater than 5, hence according to Case 2 above, we increase the digit in the tens place by one (from 8 to 9) and replace all the digits to its right by zeros giving us:

1627190

7 0

3 years ago

Read 2 more answers

Find the measure of angle x in the figure below: (1 point) Two triangles are shown such that one triangle is inverted and share

Elena L [17]

75°+75°=150°
angles in a triangle add up to 180°
180°-150°=30°
c) x=30°

5 0

2 years ago

Reduce 36/45 to lowest terms and write the numerator in the blank.

saveliy_v [14]

4/5 is your answer

Hope this helps

7 0

3 years ago

Read 2 more answers

In the nocturnal house at the zoo, there are 4 bats for every 3 spiders. If there are 112 bats and spiders, how many are bats? H

In-s [12.5K]

Answer:

9

Step-by-step explanation:

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5 0

3 years ago

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