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3 years ago

LMN≅PQR

1 answer:

7 0

X = 56°.

Based on the congruence statement, m∠L = m∠P; m∠M = m∠Q; and m∠N = m∠R.

Since m∠L = 72°, m∠P = 72°.
Since m∠N = x°, m∠R = x°.
We are given that m∠Q = 52°.

We can find the value of x by subtracting the given angles from 180:
180-(72+52) = 180-124 = 56°

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Turn 1 1/5 to improper fraction

Alex17521 [72]

Answer:

Explanation:

Step 1

Multiply the denominator by the whole number

5 × 1 = 5

Step 2

Add the answer from Step 1 to the numerator

5 + 1 = 6

Step 3

Write answer from Step 2 over the denominator

<h2><u>6/5</u></h2><h2><u></u></h2><h2><u>I hope this answer helps you out! Brainliest would be appreciated.</u></h2>

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3 years ago

mr.rivera bought some erasers to give his students. on monday , mr.rivera gave out 22 erasers. on Tuesday, he gave out 39 earser

Mumz [18]

Answer:

178 ERASERS

Step-by-step explanation:

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6 0

3 years ago

If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form

algol13

Answer:

$\frac{d^2y}{dx^2} = \frac{-4}{3}$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Algebra I</u>

Factoring

<u>Calculus</u>

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

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

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

Step-by-step explanation:

<u>Step 1: Define</u>

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

<u>Step 2: Differentiate Pt. 1</u>

<em>Find 1st Derivative</em>

Implicit Differentiation [Basic Power Rule]: $-y'-6x^2=2yy'$
[Algebra] Isolate <em>y'</em> terms: $-6x^2=2yy'+y'$
[Algebra] Factor <em>y'</em>: $-6x^2=y'(2y+1)$
[Algebra] Isolate <em>y'</em>: $\frac{-6x^2}{(2y+1)}=y'$
[Algebra] Rewrite: $y' = \frac{-6x^2}{(2y+1)}$

<u>Step 3: Differentiate Pt. 2</u>

<em>Find 2nd Derivative</em>

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}$
[Derivative] Back-Substitute <em>y'</em>: $y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}$

<u>Step 4: Find Slope at Given Point</u>

[Algebra] Substitute in <em>x</em> and <em>y</em>: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Multiply: $y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}$
[Pre-Algebra] Divide: $y''(-1,-2) = \frac{-36+24 }{9}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-12}{9}$
[Pre-Algebra] Simplify: $y''(-1,-2) = \frac{-4}{3}$