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2 years ago

What is the interval of increase and decrease of f(x)= (x-4)^2 -1

Mathematics

1 answer:

nevsk [136]2 years ago

3 0

Answer:

(−∞,4)

Step-by-step explanation:

Explanation:

To determine the interval that f(x) is increasing.

increasing when f'(x)>0f'(x)=−2(x−4)← using chain rule

solve −2(x−4)>0

⇒−2x+8>0

⇒x<4

interval is (−∞,4)

graph{-(x-4)^2-3 [-8.89, 8.89, -4.445, 4.44]}

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I dont understand and this is very urgent so can someone help me out please?

wariber [46]

Answer:

Please Find the solution below

Step-by-step explanation:

Let us say the two equations are

x+y=5 --------------(A)

x-y=1 -------------(B)

Let us solve them for x and y by adding them

2x=6

x=3

Hence from (A)

3+y=5

y=2

Hence our solution is

x=3, y=2

Adding same number to equation (A) say 2 we get

x+y+2=5+2

x+y=5+2-2

x+y=5

Hence equation remains the same while adding same number to each side.

Same thing happens if we add same number to equation (B)

Hence we draw the conclusion that the solution remains the same if same number is added to each side of the original equation.

6 0

3 years ago

Solve the following new equation

mixer [17]

Answer:

<h3>x = 1.54 or x = - 4.54 to 2 decimal places</h3>

Step-by-step explanation:

x² + 3x - 7 = 0

Using the quadratic formula

a = 1 b = 3 c = - 7

$x = \frac{3 ± \sqrt{ {3}^{2} - 4(1)( - 7) } }{2(1)} \\ x = \frac{ - 3± \sqrt{9 + 28} }{2} \\ \\ x = \frac{ -3 ± \sqrt{37} }{2} \\ \\ x = \frac{ - 3 + \sqrt{37} }{2} \: \: \: \: \: or \: \: \: x = \frac{ - 3 - \sqrt{37} }{2} \\ \\ x = 1.54 \: \: \: or \: \: x = - 4.54$

Hope this helps you

7 0

2 years ago

All of the following are true statements except _____.

shutvik [7]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

The false (wrong) equation among them is : A

$\qquad \sf \dashrightarrow \: | - 93| = - 93$

Because any number in a modulus comes out to be positive.

4 0

2 years ago

The surface area of a typical classroom floor is closest to______.

avanturin [10]

Answer would be C. Why A would be incorrect: 100m^2 is 100m × 100 m. That would be as big as a football field. Why B would be incorrect: 1cm^2 is 1cm × 1cm. You can use a regular 15 cm to measure a piece of paper with the length of 1cm each side. Why C would be correct: 1m^2 = 1m × 1m and it is also equals to 60cm × 60 cm. A 1 meter ruler = 60 cm = 4 times of your average 15 cm ruler. So, it is reasonable that a classroom is 10m by 10m in length and breadth. Why D would be incorrect: The same explanation applies to here as well. Things that can be measured with 1m × 1m is a square table so your classroom can't be that small to fit an average of 30 to 40 students in there. I hope my explanations were detailed and easy to understand. :)

4 0

2 years ago

Read 2 more answers

8/x-5 - 9/x-4 = 5/x^2-9x+20

adelina 88 [10]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2752942

_______________

Solve the equation:

$\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}$

Reduce the fractions at the left side so that they have the same denominator:

$\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}$

Numerators must be equal:

$\mathsf{8x-32-(9x-45)=5}\\\\
\mathsf{8x-32-9x+45=5}\\\\
\mathsf{8x-9x=5+32-45}\\\\
\mathsf{-x=-8}\\\\
\mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}$

I hope this helps. =)

Tags: <em>rational equation fraction solution algebra</em>

7 0

2 years ago