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4 years ago

Angles that measure 135 degrees and 45 degrees are complementary, True or false?

Mathematics

2 answers:

vagabundo [1.1K]4 years ago

7 0

Answer:

False

Step-by-step explanation:

Complementary angles are when either of two angles whose sum is 90°.

forsale [732]4 years ago

6 0

Answer:

FAlse

complementary angles measures 90

as 45 + 135 = 180 these are supplementary angles

Step-by-step explanation:

You might be interested in

Please Evaluate 9[y^3(z-6)-5] for y=2 and z=8

mafiozo [28]

I believe the answer would be 99

plug in the variables and solve to answer this question.

3 0

3 years ago

Read 2 more answers

Your classmate says that the solutions to the given inequality are x > -2. Look carefully at the work. Should your friend hav

LuckyWell [14K]

Answer:

x < -2 (No, he shouldn't have reversed the inequality symbol).

Step-by-step explanation:

Given the algebraic expression;

19 - 2(1 - x) < 13

Opening up the bracket, we have;

19 + (-2*1) + (-2*-x) < 13

19 + (-2) + (2x) < 13

19 - 2 + 2x < 13

17 + 2x < 13

Rearranging the equation, we have;

2x < 13 - 17

2x < -4

Dividing both sides by 2, we have;

x < -2

Hence, your friend shouldn't have reversed the inequality symbol (sign).

Note: you should only flip the inequality symbol (sign) when you're dividing by a negative number such as -2.

For example, if we had re-arranged the equation as;

17 - 13 < -2x

4 < -2x

Dividing both sides by -2, we have;

4 > x

6 0

3 years ago

Write an equation for each translation of y=|x|. 9 units down

kenny6666 [7]

To translate a function downwards by 9 units you simply subtract 9 from the function, because the resulting y value is reduced by 9 so the graph of y is reduced, shifted downwards by 9. So if:

y=|x| then this shifted downwards by 9 units is:

y=|x|-9

3 0

4 years ago

Help with num 1 please.

KengaRu [80]

Answer:

(i) $\displaystyle y' = (6x - 1)ln(2x + 1) + \frac{2x(3x - 1)}{2x + 1}$

(ii) $\displaystyle y' = \frac{2x}{ln(x)} - \frac{x^2 + 2}{x(lnx)^2}$

(iii) $\displaystyle y' = \frac{e^x[xln(2x) + 1]}{x}$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Exponential Differentiation

Logarithmic Differentiation

Step-by-step explanation:

(i)

Step 1: Define

Identify

$\displaystyle y = (3x^2 - x)ln(2x + 1)$

Step 2: Differentiate

Product Rule: $\displaystyle y' = (3x^2 - x)'ln(2x + 1) + (3x^2 - x)[ln(2x + 1)]'$
Basic Power Rule/Logarithmic Differentiation [Chain Rule]: $\displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)\frac{1}{2x + 1}(2x + 1)'$
Basic Power Rule: $\displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)\frac{2}{2x + 1}$
Simplify [Factor]: $\displaystyle y' = (6x - 1)ln(2x + 1) + \frac{2x(3x - 1)}{2x + 1}$

(ii)

Step 1: Define

Identify

$\displaystyle y = \frac{x^2 + 2}{lnx}$

Step 2: Differentiate

Quotient Rule: $\displaystyle y' = \frac{(x^2 + 2)'lnx - (x^2 + 2)(lnx)'}{(lnx)^2}$
Basic Power Rule/Logarithmic Differentiation: $\displaystyle y' = \frac{2xlnx - (x^2 + 2)\frac{1}{x}}{(lnx)^2}$
Rewrite: $\displaystyle y' = \frac{2xlnx}{(lnx)^2} - \frac{(x^2 + 2)\frac{1}{x}}{(lnx)^2}$
Simplify: $\displaystyle y' = \frac{2x}{ln(x)} - \frac{x^2 + 2}{x(lnx)^2}$

(iii)

Step 1: Define

Identify

$\displaystyle y = e^xln(2x)$

Step 2: Differentiate

Product Rule: $\displaystyle y' = (e^x)'ln(2x) + e^x[ln(2x)]'$
Exponential Differentiation/Logarithmic Differentiation [Chain Rule]: $\displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})(2x)'$
Basic Power Rule: $\displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})2$
Simplify: $\displaystyle y' = e^xln(2x) + \frac{e^x}{x}$
Rewrite: $\displaystyle y' = \frac{xe^xln(2x) + e^x}{x}$
Factor: $\displaystyle y' = \frac{e^x[xln(2x) + 1]}{x}$