Equivalent Expressions for 6(4t+5)+2

the graph of a line has a slope of -2/3 and a y intercept of (0,2). rewrite the equation in standard for (Ax+By=C) with a positi

sergiy2304 [10]

Answer:

2x + 3y = 6

Step-by-step explanation:

obtain the equation in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

here m = - $\frac{2}{3}$ and c = 2

y = - $\frac{2}{3}$ x + 2 ← in slope-intercept form

multiply all terms by 3 to eliminate the fraction

3y = - 2x + 6 ( add 2x to both sides )

2x + 3y = 6 ← in standard form

8 0

3 years ago

Sea un cuadrado de 2 pulgadas de lado uniendo los puntos medios se obtiene otro cuadrado inscrito en el anterior si repetimos es

Ne4ueva [31]

Answer:

1) La serie geométrica formada es

4, 2, 1,..., ∞

2) La suma al infinito de las áreas de los cuadrados es 8 in.²

Step-by-step explanation:

1) El área del primer cuadrado, a₁ = 2² = 4 pulgadas²

El área del siguiente cuadrado, a₂ = (√ (1² + 1²)) ² = (√2) ² = 2 pulg²

El área del siguiente cuadrado, a₃ = ((√ (2) / 2) ² + (√ (2) / 2) ²) = 1 pulg²

Por lo tanto, la razón común, r = a₂ / a₁ = 2/4 = a₃ / a₂ = 1/2

Las áreas de los cuadrados progresivos forman una progresión geométrica como sigue;

4, 4×(1/2), 4 ×(1/2)²,...,4× $(1/2)^{\infty}$

De donde obtenemos la serie geométrica formada de la siguiente manera;

4, 2, 1,..., ∞

2) La suma de 'n' términos de una progresión geométrica hasta el infinito para -1 <r <1 se da como sigue;

$S_{\infty} = \dfrac{a}{1 - r}$

Por lo tanto, la suma de las áreas de los cuadrados hasta el infinito se obtiene sustituyendo los valores de 'a' y 'r' en la ecuación anterior de la siguiente manera;

$La \ suma \ al \ infinito \ del \ cuadrado \ S_{\infty} = \dfrac{4 \ in.^2}{1 - \dfrac{1}{2} } = \dfrac{4 \ in.^2}{\left(\dfrac{1}{2} \right)} = 2 \times 4 \ in.^2= 8 \ in.^2$

La suma al infinito de las áreas de los cuadrados, $S_{\infty}$ = 8 in.²

7 0

3 years ago

A city planner designs a park that is a quadrilateral with vertices at J(-3,1), K(1,3), L(5,-1), and M(-1,-3). There is an entra

Ksivusya [100]

The Quadrilateral is JKLM,

let $M_{JK}, M_{KL}, M_{LM}, M_{JM},$ be the midpoints of JK, KL, LM and JM respectively.

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Given any 2 point P(m,n) and Q(k,l),<span>

the coordinates of the midpoint of the line segment PQ are given by the formula:

$M_{PQ}=( \frac{m+k}{2} ,
\frac{n+l}{2})$ , </span>

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thus the coordinates of points $M_{JK}, M_{KL}, M_{LM}, M_{JM},$

are as follows:

$M_{JK}= (\frac{-3+1}{2}, \frac{1+3}{2})=(-1,2), \\\\M_{KL}= (\frac{1+5}{2}, \frac{3-1}{2}=(3,1), \\\\M_{LM}= (\frac{5-1}{2}, \frac{-1-3}{2})=(2, -2),\\\\ M_{JM}= (\frac{-3-1}{2}, \frac{1-3}{2})=(-2,-1)$

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The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

$|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}
}$ </span>

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thus the distances connecting the opposite entrances can be calculated as follows:

$|M_{JK},M_{LM}|= \sqrt{ (-1-2)^{2} + (2-(-2))^{2} }= \sqrt{9+16}=5$

$|M_{KL}M_{JM}|= \sqrt{ (3-(-2))^{2} + (1-(-1))^{2}}= \sqrt{25+4}= \sqrt{29}=5.39$

Thus the total distance of the paths joining the opposite entrances is

5+5.39 units = 50 m + 53.9 m = 104 m (rounded to the nearest meter)

Answer: 104 m

8 0

3 years ago

How can you use properties to solve equations with variables on both sides

bija089 [108]

You can use properties to solve equations with variables on both side by simplifying get the variable on one side. solve using inverse operations then check to see if it fits in right .

5 0

3 years ago

The line which passes through -2,5 and 6,8

babunello [35]

Answer:

y=(3/8)x+5.75

Step-by-step explanation:

y=mx+b

m=slope

b=y-intercept

m=(8-5)/(6+2)=3/8

y=mx+b

5=3/8(-2)+b

5=-3/4+b

b=5.75

y=mx+b

y=(3/8)x+5.75

3 0

3 years ago