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3 years ago

Show me how to get this answer show your work

Mathematics

2 answers:

icang [17]3 years ago

8 0

The measure of angle MKJ is equal to the sum of angles MKL and JKL by the angle addition postulate. that means
2x+10 + 3x-5=80. solve for x...

5x+5=80
5x=75
x=15

now plug 15 in n for x in both angles. you'll see that the measure of both angles is 40. since segment KL cuts the big angle exactly in half, it is the angle bisector. your checked answer is the right one.

Rama09 [41]3 years ago

6 0

We know that angle MKJ is comprised of angle MKL and angle LKJ. That means if we add MKL and LKJ, we should get 80 degrees, which is the measure of angle MKJ.

$MKL + LKJ= MKJ \implies \\ 2x+10+3x-5=80 \implies \\ 5x+5=80 \implies\\ 5x=75 \implies \\ x=15$

So, we know that our x is 15. That is not enough to tell whether KL is an angle bisector, because we have to evaluate both MKL and LKJ with x=15, so:

$MKL=2(15)+10 = 40\\ LKJ=3*15-5=40$

So we see that these two angles are actually bisectors, and the third question best describes this phenomenon.

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Answer:

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Step-by-step explanation:

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3 years ago

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Soloha48 [4]

No x-intercept/zero

Step by step

To find x-intercept/zero, substitute f(x)=0

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The statement is false for any value of

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Your welcome:)

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3 years ago

If f(x)=×^2-81andg(x)

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Answer:

the correct choice is marked

Step-by-step explanation:

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3 years ago

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nataly862011 [7]

Answer:

$\displaystyle y' = 2x + 3\sqrt{x} + 1$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Terms/Coefficients
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Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = (x + \sqrt{x})^2$

Step 2: Differentiate

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Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = 2(x + x^{\frac{1}{2}})^{2 - 1} \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}]$
Simplify: $\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}]$
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Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

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