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olchik [2.2K]
3 years ago
11

Mario compared the slope of the function graphed below to the slope of the linear function that has an x-intercept of 1 and a y-

intercept of -2

Mathematics
1 answer:
Lelu [443]3 years ago
7 0

Answer:

The slope of the function graphed is less than the linear function that has an x-intercept of 1 and a y-intercept of -2.

Step-by-step explanation:

It is given that x-intercept is 1 and y-intercept is -2.

Now, the coordinate of the point is (1,0) and (0,-2).

We know that,

Slope = \dfrac{y_2-y_1}{x_2-x_1}

S_1=\dfrac{-2-0}{0-1}

S_1 = \dfrac{-2}{-1}

S_1=2

Now, Slope of the graph given

Co-ordinate is (-5,0) and (1,2).

Using the formula of slope

S_2 = \dfrac{2-0}{1-(-1)}

S_2 = \dfrac{2}{1+1}

S_2 = \dfrac{2}{2}

S_2 =1

Now, S_1>S_2

Hence, the slope of function graphed is less than linear function that has an x-intercept of 1 and a y-intercept of -2.

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Please help solve the equation and please don’t take advantage of the points.
vfiekz [6]

Answer:

m=2/3, b=4

Step-by-step explanation:

Using the rise over run method, you can count from two points. Also since this line is positive. The b is found by seeing where the line crosses the y axis. It crosses at 4.

5 0
3 years ago
Al factorizar el trinomio cuadrado perfecto, obtenemos el siguiente resultado: (que no se como resolver) xd algun pro que sepa r
PolarNik [594]

Answer:

\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8=\left(\frac{10}{9}m^4p^{6}q^{8}z-\frac{1}{7}mpz^4\right)^2

Step-by-step explanation:

<u>Trinomio Cuadrado Perfecto</u>

El producto notable llamado cuadrado de un binomio se expresa como:

(a-b)^2=a^2-2ab+b^2

Si se tiene un trinomio, es posible convertirlo en un cuadrado perfecto si cumple con las condiciones impuestas en la fórmula:

* El primer término es un cuadrado perfecto

* El último término es un cuadrado perfecto

* El segundo término es el doble del proudcto de los dos términos del binomio.

Tenemos la expresión:

\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8

Calculamos el valor de a como la raiz cuadrada del primer término del trinomio:

\displaystyle a=\sqrt{\frac{100}{81}m^8p^{12}q^{16}z^2}

\displaystyle a=\frac{10}{9}m^4p^{6}q^{8}z

Calculamos el valor de a como la raiz cuadrada del primer término del trinomio:

\displaystyle b=\sqrt{\frac{1}{49}m^2p^2z^8

\displaystyle b=\frac{1}{7}mpz^4

Nos cercioramos de que el término central es 2ab:

\displaystyle 2ab=2\frac{10}{9}m^4p^{6}q^{8}z\frac{1}{7}mpz^4

Operando:

\displaystyle 2ab=\frac{20}{63}m^5p^7q^8z^5

Una vez verificado, ahora podemos decir que:

\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8=\left(\frac{10}{9}m^4p^{6}q^{8}z-\frac{1}{7}mpz^4\right)^2

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Answers please?? Thank you!
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A= 1,3
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