Which of the following expressions has the greatest value when x=5? show how you arrived at your choice.2x^2+7

x^3-5/3

10x -2/ x-3

Answers

Answer 1
Answer:

Answer:

2x^2+7

Step-by-step explanation:

Let x=5

2x^2+7

2 ( 5)^2 +7 = 2*25 +7 = 50+7 = 57

(x^3-5)/3

(5^3 -5)/3 = (125-5)/3 = 120/3 = 40

(10x -2)/ (x-3)

(10*5-2)/(5-3) = (50-2)/(2) = 48/2 = 24


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Lena is on a two week bicycle trip. After 5 days she had ridden 212 miles. Express Lena’s rate as a unit rate

Answers

Answer:

42.4 miles per day

Step-by-step explanation:

212 miles ÷ 5 (days) = 42.4 miles

which means Lena can go at the rate of

42.4 miles per day

suzi starts her hike at a elevation if -225 below sea level. When she reaches the end of the hike, she is still nelow sea level at -127 feet. What was the vhange in elevation from the beginning of Suzi's hike to the end of the hike?

Answers

The change is elevation is 98 feet.

There are a couple different ways to get this answer. You could:
A. Subtract 127 from (positive) 225, to get the height difference. Or...
B. Subtract the starting height from the ending height (-127 - (-225)) which would also give you the height difference.

I hope this helps!

How to simply the radical

Answers

6√2 + 4 √32
try to get the same number "2" remaining under the root sign by finding what # times 2 gives 32
6√2 + 4√2×√16
since 16 is a perfect Square √16 gives 4
therefore.
6√2 + 4 ×4 √2
6√2 +16√2
these are Like terms so add them
which would give u
22√2
hoped i helped, feel free to ask any question XOX

A study is under way in Yosemite National Forest to determine that adult height of American pine trees. Specifically, the study is attempting to determine what factors aid a tree in reaching heights greater than 60 feet tall. It is estimated that the forest contains 25,000 adult American pines. The study involves collecting heights from 250 randomly selected adult American pine trees and analyzing the results. Identify the population from which the study was sampled

Answers

Answer:

The 25,000 adult American pine trees in the forest.

Step-by-step explanation:

Consider the provided information.

It is given that the forest contains 25,000 adult American pines. The study involves collecting heights from 250 randomly selected adult American pine trees and analyzing the results.

A population is the set of data which includes all of the elements. If one or more observations are drawn form the population then it called a sample.

Here the population is 25,000 adult American pines.

Thus, the required answer is: The 25,000 adult American pine trees in the forest.

When a number is increased by 26, the result is tripled. Then the result is increased by 72. If the final result is 1/2 of the number, what is the value of this number?

Answers

Answer:

-60.

Step-by-step explanation:

Let the unknown number be x.

Number is increased by 26 = x+26

Then result is tripled = 3(x+26)

Then the result is increased by 72 = 3(x+26)+72

Final result is (1)/(2) of the number = (1)/(2)x

3(x+26)+72=(x)/(2)

3x+78+72=(x)/(2)

3x+150=(x)/(2)

Isolate variable terms.

3x-(x)/(2)=-150

(6x-x)/(2)=-150

Multiply both sides by 2.

5x=-300

Divide both sides by 5.

x=-(300)/(5)

x=-60

Therefore, the required number is -60.

Pleeease open the image and hellllp me

Answers

1. Rewrite the expression in terms of logarithms:

y=x^x=e^(\ln x^x)=e^(x\ln x)

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of y is denoted y' )

y'=e^(x\ln x)(x\ln x)'=x^x(x\ln x)'

y'=x^x(x'\ln x+x(\ln x)')

y'=x^x\left(\ln x+\frac xx\right)

y'=x^x(\ln x+1)

2. Chain rule:

y=\ln(\csc(3x))

y'=\frac1{\csc(3x)}(\csc(3x))'

y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)

y'=-3\sin(3x)\cot^2(3x)

Since \cot x=(\cos x)/(\sin x), we can cancel one factor of sine:

y'=-3(\cos^2(3x))/(\sin(3x))=-3\cos(3x)\cot(3x)

3. Chain rule:

y=e^{e^(\sin x)}

y'=e^{e^(\sin x)}\left(e^(\sin x)\right)'

y'=e^{e^(\sin x)}e^(\sin x)(\sin x)'

y'=e^{e^(\sin x)+\sin x}\cos x

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to e, you can use the change-of-base formula first:

\log_2x=(\ln x)/(\ln2)

Then

(\log_2x)'=\left((\ln x)/(\ln 2)\right)'=\frac1{\ln 2}

So we have

y=\cos^2(\log_2x)

y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'

y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'

y'=-\frac2{\ln2}\cos(\log_2x)\sin(\log_2x)

and we can use the double angle identity and logarithm properties to condense this result:

y'=-\frac1{\ln2}\sin(2\log_2x)=-\frac1{\ln2}\sin(\log_2x^2)

5. Differentiate both sides:

\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'

2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\frac{xy'}y+\ln y=0

-\left(2y-\sin x\,e^y-\frac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)

y'=(2x+\cos x\,e^y\ln y)/(2y-\sin x\,e^y-\frac xy)

y'=(2xy+\cos x\,ye^y\ln y)/(2y^2-\sin x\,ye^y-x)

6. Same as with (5):

\left(\sin(x^2+\tan y)+e^(x^3\sec y)+2x-y+2\right)'=0'

\cos(x^2+\tan y)(x^2+\tan y)'+e^(x^3\sec y)(x^3\sec y)'+2-y'=0

\cos(x^2+\tan y)(2x+\sec^2y y')+e^(x^3\sec y)(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0

\cos(x^2+\tan y)(2x+\sec^2y y')+e^(x^3\sec y)(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0

\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^(x^3\sec y)-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^(x^3\sec y)+2\right)

y'=-(2x\cos(x^2+\tan y)+3x^2\sec y\,e^(x^3\sec y)+2)/(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^(x^3\sec y)-1)

7. Looks like

y=x^2-e^(2x)

Compute the second derivative:

y'=2x-2e^(2x)

y''=2-4e^(2x)

Set this equal to 0 and solve for x :

2-4e^(2x)=0

4e^(2x)=2

e^(2x)=\frac12

2x=\ln\frac12=-\ln2

x=-\frac{\ln2}2