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3 years ago

Help me please I need help

Mathematics

1 answer:

Lemur [1.5K]3 years ago

4 0

It actualy gives you the minimum of the function
it helps you because fonding the minimum y will give you the range

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<img src="https://tex.z-dn.net/?f=%20x%20-%202%5Csqrt%7Bx%20-%203%20%3D%200%7D%20" id="TexFormula1" title=" x - 2\sqrt{x - 3 = 0

GrogVix [38]

Answer:

No Real Solutions

Step-by-step explanation:

So we have:

$x-2\sqrt{x-3}=0$

First, determine the domain restrictions. The expression under the radical cannot be less than 0. Therefore:

$x-3\geq 0\\x\geq 3$

Therefore, our final answers must be greater than or equal to 3:

Now, going back to the original equation, subtract x from both sides:

$-2\sqrt{x-3}=-x$

Now, square both sides:

$4(x-3)=x^2$

Distribute:

$4x-12=x^2$

Subtract 4x and add 12 to both sides:

$0=x^2-4x+12$

This isn't factor-able. Let's use the quadratic formula. a is 1, b is -4 and c is 12:

$x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substitute:

$x=\frac{4\pm\sqrt{(-4)^2-4(1)(12)}}{2(1)}$

Simplify the radical:

$x=\frac{4\pm\sqrt{-32}}{2(1)}$

The number under the radical is negative. In other words, there are no real solutions.

7 0

2 years ago

The circumference of a circle is 32 centimeters. Which is closest to the radius? A. 10.2 centimeters B. 5.1 centimeters C. 16 ce

pshichka [43]

The answer is B 5.1 cm

8 0

2 years ago

Read 2 more answers

Read the excerpt from The Land.

babunello [35]

From the conversation it can be inferred that Paul's approach to getting on the horse is methodical (option c).

<h3>What is methodical behavior?</h3>

A methodical attitude or behavior refers to the preference of people to use an organized method and sequences to perform an activity correctly.

According to the above, it can be inferred that Paul behaves in a methodical manner because he prefers to relate to the horse slowly and in a controlled manner before proceeding to mount it.

Learn more about methodical in: brainly.com/question/14560322

#SPJ1

5 0

1 year ago

Help!!<br><br> which expression is equivalent to (m^5 n /pq^2) ^4

lana66690 [7]

$\text{Use:}\\\\\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(a^n)^m=a^{nm}\\\\......................................................\\\\\left(\dfrac{m^5n}{pq^2}\right)^4=\dfrac{(m^5n)^4}{(pq^2)^4}=\dfrac{(m^5)^4n^4}{p^4(q^2)^4}=\dfrac{m^{20}n^4}{p^4q^8}$

8 0

3 years ago

Will give brainliest please help asap

ella [17]

What do you need help with?

4 0

2 years ago