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3 years ago

BIVA

Mathematics

1 answer:

sladkih [1.3K]3 years ago

8 0

Answer:

OPEN PHONE

Step-by-step explanation:

bhad m jaaa sun pagal

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Well the answer is either Jesus or applesauce

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A number cube is rolled 450 times and the

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Answer:

A.) 0.33

Step-by-step explanation:

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3 years ago

Find the distance between the two points in simplest radical form. (7,2) and (-1, -3)

Paul [167]

Answer:

The answer is

<h2> $\sqrt{89} \: \: \: \: units \\$ </h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x_1 - x_2})^{2} + ({y_1 - y_2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(7,2) and (-1, -3)

The distance between them is

$d = \sqrt{( {7 + 1})^{2} + ({2 + 3})^{2} } \\ = \sqrt{ {8}^{2} + {5}^{2} } \\ = \sqrt{64 + 25} \\ = \sqrt{89} \: \: \: \: \: \: \: \: \: \:$

We have the final answer as

$\sqrt{89} \: \: \: \: units \\$

Hope this helps you

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4 years ago

Hailey, Carla and Britney had a contest

Julli [10]

Answer:

Hailey won by holding her breath for 102 seconds

Step-by-step explanation:

Britney held her breath for 44 seconds

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3 years ago

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Y=sqrt(x)(8x-5) find the derivative

garri49 [273]

Answer:

$\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$

General Formulas and Concepts:

Algebra I

Exponentials [Fractions] - Are radicals
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \sqrt{x}(8x - 5)$

Step 2: Differentiate

$\displaystyle f(x) = \sqrt{x}, \ g(x) = (8x - 5)$

Product Rule: $\displaystyle y' = \frac{d}{dx}[\sqrt{x}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$
Rewrite: $\displaystyle y' = \frac{d}{dx}[x^{\frac{1}{2}}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{1 - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{0}$
Rewrite: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8$
Multiply: $\displaystyle y' = \frac{8x + 5}{2x^{\frac{1}{2}}} + 8\sqrt{x}$
Rewrite: $\displaystyle y' = \frac{8x + 5}{2\sqrt{x}} + 8\sqrt{x}$
Add/Rewrite: $\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$

3 0

3 years ago