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Alina [70]
4 years ago
15

⚠️ Show work for points ⚠️

Mathematics
1 answer:
Nikitich [7]4 years ago
4 0

Answer:

D

Step-by-step explanation:

A is not, because 2-5x = -5x+2

(nothing has changed but the order)

B is not, because 3-5+2x = 2x-2

because 3-5 = -2. so -2+2x can be rewritten as 2x-2.

C is not, because 5x-2x = 3x

because you combine like terms, and 5-2 is 3

D <em>i</em><em>s</em><em>,</em> because 2x-2+3x = 5x-2

because you combine like terms (2x+3x=5x) and -2 is left.

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Examine the diagram. Triangle A B C. Side A C is parallel to a line. The line is at angle B. The angles formed by the line are 7
Marina CMI [18]

Answer:

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8 0
4 years ago
Shane and Karen want to measure the length of a soccer field. Should they use centimeters or meters to measure it? Explain.
Ber [7]

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3 0
3 years ago
Read 2 more answers
En un aeropuerto dos aviones A1 y A1 se acercan para aterrizar, si la ecuación de la trayectoria del primero es -x+2y-100=0, mie
liraira [26]

No hay solución en la que ambas trayectorias estén en la misma posición, puesto que no existe el riesgo de que sufran un choque entre ellos.

-------------------------

La trayectoria del primero es:

-x + 2y - 100 = 0

x_1 = 2y - 100

Para el segundo, tiene-se que:

-x + 200 + 2y = 0

x_2 = 2y + 200

Igualando los valores de x:

x_1 = x_2

2y - 100 = 2y + 200

0y = 300

No hay solución en la que ambas trayectorias estén en la misma posición, puesto que no existe el riesgo de que sufran un choque entre ellos.

Un problema similar es dado en brainly.com/question/24653364

3 0
3 years ago
What value of b will cause the system to have an infinite number of solutions?
irga5000 [103]

b must be equal to -6  for infinitely many solutions for system of equations y = 6x + b and -3 x+\frac{1}{2} y=-3

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}

Let us bring the equations in same form for sake of simplicity in comparison

\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}

Now we have two equations  

\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}

Let us first see what is requirement for system of equations have an infinite number of solutions

If  a_{1} x+b_{1} y+c_{1}=0 and a_{2} x+b_{2} y+c_{2}=0 are two equation  

\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} then the given system of equation has no infinitely many solutions.

In our case,

\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}

 As for infinitely many solutions \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}

Hence b must be equal to -6 for infinitely many solutions for system of equations y = 6x + b and  -3 x+\frac{1}{2} y=-3

8 0
3 years ago
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