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4 years ago

What is the inverse of f(x)=x^-9

1 answer:

3 0

$y = \frac{1}{x^9}\\yx^9 = 1\\x^9 = \frac{1}{y}\\x = \sqrt[9]{\frac{1}{y}}\\Therefore, \ f^{-1}(x)=\sqrt[9]{\frac{1}{x}}.$

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Please help it is about line segments?! I’ll give brainlist

shutvik [7]

Answer:

Step-by-step explanation:

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7 0

2 years ago

Eighty-seven decreased by three times a number is greater than one hundred sixty-five

fenix001 [56]

This is the concept of algebra, we are required to find a number,x, such that 87 decreased by three times a number is great than one hundred and sixty-five.

Supposed the number is x, three times the number will be:
3*x=3x
thus;
87-3x≥165
-3x≥165-87
-3x≥78
dividing both sides by -3 we get:
(-3x)/(-3x)≥78/(-3)
x≤-26
This implies that the number is less than or equal to 26

5 0

3 years ago

Is (2, -3) a solution of x + 5y = 4?

givi [52]

So for this, plug in 2 into the x variable and -3 into the y variable and solve.

$2+5(-3)=4\\ 2+(-15)=4\\ -13=4$

Since -13 = 4 is a false statement, (2,-3) is not a solution to this equation.

8 0

3 years ago

The product of a number x and 16 is equal to 7. Which of the following is the value of X

blondinia [14]

Answer:

$x = \frac{7}{16}$

Step-by-step explanation:

$x\cdot16 = 7\\\\16x=7\\\\x=\frac{7}{16}$

5 0

3 years ago

A rectangular garden must have a perimeter of 145 feet and an area of at least 1100 square feet. Describe the possible lengths o

marta [7]

Answer:

The approximate length of the garden is at least 22 feet and at most 51 feet.

Step-by-step explanation:

A rectangular garden must have a perimeter of 145 feet.

Let x feet be the lengths of the garden. Then the width of the garden can be found using the perimeter:

$P=2(\text{Length}+\text{Width})\\ \\145=2(x+\text{Width})\\ \\\text{Width}=72.5-x$

The area of the garden is

$A=\text{Length}\cdot \text{Width}\\ \\A=x\cdot (72.5-x)$

The area must be at least 1,100 square feet, then

$x(72.5-x)\ge 1,110\\ \\72.5x-x^2-1,100\ge 0\\ \\x^2-72.5x+1,100\le 0$

Solve this quadratic inequality:

$D=(-72.5)^2-4\cdot 1,100=856.25\\ \\\sqrt{D}=25\sqrt{1.37}\\ \\x_1=\dfrac{72.5-25\sqrt{1.37}}{2}\approx 22\\ \\x_2=\dfrac{72.5+25\sqrt{1.37}}{2}\approx 51\\ \\(x-22)(x-51)\le 0\\ \\x\in [22,51]$

3 0

3 years ago