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3 years ago

Solve using the square root method X2=25

Mathematics

2 answers:

aleksandrvk [35]3 years ago

8 0

Answer:

Square root =

remember :

$\boxed {x^2 = \pm a}$

x² = ± a

x = √±a

x = √±25

x = ±5

• x = 5 and -5

geniusboy [140]3 years ago

3 0

X^2=25
sqrt x^2= sqrt 25
x=5

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3 years ago

Pic attached solve for x

nordsb [41]

The answer to your question is x = -3

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3 years ago

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S_A_V [24]

Answer:

Step-by-step explanation:

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2 years ago

Water drips into a circular puddle such that the radius of the puddle, in centimeters, at time t, in seconds, is given by the eq

bulgar [2K]

Part A

Given that the puddle is circular in shape and that the radius of the puddle, in centimeters, at time t, in seconds, is given by the equation $r(t)=\sqrt{t}$ .

Then the area of the puddle is given by the area of a circle = $Area=\pi r^2$
But, given that $r(t)=\sqrt{t}$ , then $A(t)=\pi (r(t))^2=\pi(\sqrt{t})^2=\pi t$

Therefore, the equation for the area of the puddle as a function of t is given by $A(t)=\pi t$

Part B

The average rate of change of a function f(x) between x = a and x = b is given by $\frac{f(b)-f(a)}{b-a}$ .

Thus, the average rate of change of the area of the puddle with respect to time between t = 0 and t = 16 is given by $\frac{A(16)-A(0)}{16-0} = \frac{16\pi-0}{16} = \frac{16\pi}{16} =\pi$

Therefore, the average rate of change of the area of the puddle with respect to time between t = 0 and t = 16 is π.

Part C

The area of the puddle with respect to the radius is given by $A(r)=\pi r^2$

Given that $r(t)=\sqrt{t}$ , thus when t = 0, $r(0)=\sqrt{0}=0$ and when t = 16, $r(16)=\sqrt{16}=4$

Thus, the average rate of change of the area of the puddle with respect to the radius between r = 0 and r = 4 is given by

$\frac{A(4)-A(0)}{4-0} = \frac{\pi(4)^2-\pi(0)^2}{4} = \frac{16\pi}{4} =4\pi$

Therefore, the average rate of change of the area of the puddle with respect to the radius between t = 0 and t = 16 is 4π.

Part D

The circumference of a circle is given by $C=2\pi r$

Thus, the radius of the puddle in terms of circumference is given by $r= \frac{C}{2\pi}$

Thus, the area of the puddle with respect to the circumference, C, of the puddle is given by $A(C)=\pi\left( \frac{C}{2\pi} \right)^2= \frac{1}{4\pi} C^2$

Since, $C=2\pi r$ and $r(t)= \sqrt{t}$ , thus when t = 0, r = 0 and C = 0; when t = 16, r = 4 and C = 8π.

Thus the area of the puddle with respect to the circumference, C, of the puddle between C = 0 and C = 8π is given by $\frac{A(8\pi)-A(0)}{8\pi-0} = \frac{ \frac{(8\pi)^2}{4\pi}- \frac{(0)^2}{4\pi} }{8\pi} = \frac{ \frac{64\pi^2}{4\pi} }{8\pi} = \frac{16\pi}{8\pi} =2$

Therefore, the average rate of change of the area of the puddle with respect to the circumference of the puddle between t = 0 and t = 16 is 2.

5 0

3 years ago

Help me answer this question please

arlik [135]

Answer:

f'(2) = 8

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Calculus

The definition of a derivative is "the slope of the tangent line".

Derivatives of constants are 0.

Basic Power Rule:

f(x) = cxⁿ

f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

f(x) = 3x² - 4x + 2

Step 2: Find 1st Derivative

Basic Power Rule: f'(x) = 2·3x²⁻¹ - 1·4x¹⁻¹
Simplify: f'(x) = 6x - 4

Step 3: Find tangent slope

Define: f'(x) = 6x - 4, x = 2
Substitute: f'(2) = 6(2) - 4
Multiply: f'(2) = 12 - 4
Subtract: f'(2) = 8

Step 4: Identify

f'(2) = 8 tells us that at x = 2, the slope of the tangent line is 8.

7 0

3 years ago

Solve using the square root method X2=25​

Solve using the square root method X2=25