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3 years ago

Determine whether the function has a maximum or minimum value. f(×)= x^2+9

2 answers:

3 0

$f(x)= x^2+9\\\\for\ each\ x\in R\\\\x^2 \geq 0\ \ \ \Rightarrow\ \ \ x^2+9 \geq 9\ \ \ \Rightarrow\ \ \ f(x) \geq 9\\\\y_{min}=9\\ y_{max}\ \ \ \rightarrow\ \ \ not\ exist$

miv72 [106K]3 years ago

3 0

$a>0$ so the function has a minimum, but no maximum.

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The area of Gateway National Recreation is about twenty-six thousand, six hundred and sixty-three hundredths. Which shows this n

Delicious77 [7]

Answer:

B. 26,606.63

Step-by-step explanation:

The area of Gateway National Recreation Area is about twenty-six thousand, six hundred six and sixty-three hundredths. Which shows this number in standard form? A. 26,606.3 B. 26,606.63

C. 26,660.63 D. 26,666.3

Solution

Break down and writing in numerals of twenty-six thousand, six hundred six and sixty-three hundredths.

Twenty six thousand = 26,000

Six hundred six = 606

Sixty - three hundredth = 0.63

Adding

26,000 + 606 + 0.63

We have

26,606.63

The correct answer is B

5 0

3 years ago

William bought a new sweater at the store when they were having a 45% off sale.

Lady bird [3.3K]

William payed $8.55 at the discount price

5 0

3 years ago

Read 2 more answers

Anyone can help me with my math homework please?

Vedmedyk [2.9K]

Answer:

Step-by-step explanation:

hello,

so we know y in terms of t and x in terms of t and we need to find y in terms of x

$x=21t^2\sqrt{x}=\sqrt{21}*t \ \ as \ \ t>=0 \ \ So\\t=\sqrt{\dfrac{x}{21}}$

and then

$y=f(x)=3\sqrt{\dfrac{x}{21}}+5=\sqrt{\dfrac{9x}{21}}+5=\sqrt{\dfrac{3x}{7}}+5$

hope this helps

4 0

3 years ago

Solve this x²+8/10 -x = 10/2

AysviL [449]

$x^2+\frac{8}{10}-x=\frac{10}{2}\\
x^2-x+\frac{8}{10}-\frac{10}{2}=0\\
x^2-x+\frac{8}{10}-\frac{50}{10}=0\\
x^2-x-\frac{42}{10}=0\\
x^2-x+\frac{1}{4}-\frac{1}{4}-\frac{42}{10}=0\\
(x-\frac{1}{2})^2=\frac{1}{4}+\frac{42}{10}\\
(x-\frac{1}{2})^2=\frac{5}{20}+\frac{84}{20}\\
(x-\frac{1}{2})^2=\frac{89}{20}\\
x-\frac{1}{2}=\sqrt{\frac{89}{20}} \vee x-\frac{1}{2}=-\sqrt{\frac{89}{20}}\\
x=\frac{1}{2}+\frac{\sqrt{89}}{\sqrt{20}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{\sqrt{20}}\\


$
$x=\frac{1}{2}+\frac{\sqrt{89}}{2\sqrt{5}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{2\sqrt5}}\\
x=\frac{1}{2}+\frac{\sqrt{445}}{10} \vee x=\frac{1}{2}-\frac{\sqrt{445}}{10}}\\
x=\frac{5+\sqrt{445}}{10} \vee x=\frac{5-\sqrt{445}}{10}}\\$

7 0

3 years ago

Which is the area of a square with one side that measures 1 foot? A. 1 square inch B. 100 square inches C. 144 square inches D.

cluponka [151]

Answer:

Option C. 144 square inches is the correct answer.

Step-by-step explanation:

We have been given a square with one side with a measurement of 1 foot and we have to calculate the area in square inches.

First we know that 1 foot of length = 12 inches on scale

Therefore side of the square is 12 inches

Now the area of the square will be = (Side×Side) = (Side)²

= (12)² = 144 square inches.

So the area of the square will be 144 square inches.

6 0

3 years ago