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3 years ago

If x-5=2 then find the value of 6x-20

2 answers:

7 0

First solve the first equation to find x.

x-5=2
x=7

Now plug it in to the other equation to get your answer.

6(7)-20
42-20
22

The answer is 22.

swat323 years ago

4 0

So x in the fist problem is 7. So the x is substituted be 7. guessing you want me to solve,
6x-20
=6(7)-20
=42-20
=22

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1 Simplify the following expressions: 1) -4(a + 2) - a

aleksandrvk [35]

1.C) -4a-8-a = -5a-8
2. C)-2 - x - 4.3= -6.3 -x
3.A) -2x+6-x-4= -3x+2
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8 0

3 years ago

Read 2 more answers

Translate the expression: “the difference between twice a number and 23”

kherson [118]

Answer:

23- 2x

Step-by-step explanation:

Twice a number is 2 and "x" represents "a number"

Key words: Twice, Difference (subtraction)

7 0

3 years ago

Find the derivative of following function.

Aleks04 [339]

Answer:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}$

General Formulas and Concepts:
Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]:
$\displaystyle (cu)' = cu'$

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
$\displaystyle (uv)' = u'v + uv'$

Derivative Rule [Quotient Rule]:
$\displaystyle \bigg( \frac{u}{v} \bigg)' = \frac{vu' - uv'}{v^2}$

Derivative Rule [Chain Rule]:
$\displaystyle [u(v)]' = u'(v)v'$

Step-by-step explanation:

*Note:

Since the answering box is way too small for this problem, there will be limited explanation.

Step 1: Define

Identify.

$\displaystyle y = \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \times \frac{\tan^2 x + 5x}{\csc^2 x + 3}$

Step 2: Differentiate

We can differentiate this function with the use of the given derivative rules and properties.

Applying Product Rule:

$\displaystyle y' = \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) '$

Differentiating the first portion using Quotient Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( \cos^2 x - 3\sqrt{x} + 6 \big)' \big( \sin^2 x + 6 \big) - \big( \sin^2 x + 6 \big)' \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Differentiating the second portion using Quotient Rule again:

$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( \tan^2 x + 5x \big)' \big( \csc^2 x + 3 \big) - \big( \csc^2 x + 3 \big)' \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule again:
$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Substitute in derivatives:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2} \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Simplify:

$\displaystyle y' = \frac{\big( \tan^2 x + 5x \big) \bigg[ \big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - 2 \sin x \cos x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \bigg]}{\big( \sin^2 x + 6 \big)^2 \big( \csc^2 x + 3 \big)} + \frac{\big( \cos^2 x - 3\sqrt{x} +6 \big) \bigg[ \big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) + 2 \csc^2 x \cot x \big( \tan^2 x + 5x \big) \bigg] }{\big( \csc^2 x + 3 \big)^2 \big( \sin^2 x + 6 \big)}$