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3 years ago

HELPPPP !! Find the measure of angle LMN if the measure of angle LMP is 107 !!

Mathematics

1 answer:

Rama09 [41]3 years ago

5 0

Answer:

m∠LMN = 29°

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Step-by-step explanation:

Step 1: Set up equation

m∠LMN and m∠NMP must add up to m∠LMP

(2x - 1)° + (5x + 3)° = 107°

Step 2: Solve for x

Combine like terms: 7x + 2 = 107
Isolate x term: 7x = 105
Isolate x: x = 15

Step 3: Find m∠LMN

Define: m∠LMN = (2x - 1)°
Substitute in x: m∠LMN = (2(15) - 1)°
Multiply: m∠LMN = (30 - 1)°
Subtract: m∠LMN = 29°

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$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

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$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

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the ratio of large triangles to small triangles is 2-6 .If there are 100 triangles in the design ,how many small triangles are t

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Answer:

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Step-by-step explanation:

1. Reduce the ratio: 2-6 = 1-3

2. Write each of them as a fraction of a whole:

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Please help me 33/2 + 3y/5 = 7y/10 + 15

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We are given with an equation in variable y and we need to solve for y . So , now let's start !!!

We are given with ;

${:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}$

Take LCM on both sides :

${:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}$

Multiplying both sides by 10 ;

${:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}$

${:\implies \quad \sf 165+6y=7y+150}$

Can be further written as ;

${:\implies \quad \sf 7y+150=165+6y}$

Transposing 6y to LHS and 150 to RHS

${:\implies \quad \sf 7y-6y=165-150}$

${:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}$

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