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2 years ago

YES OR NO QUESTION PLEASE HELP ILL GIVE BRAINLIEST

Mathematics

1 answer:

Viefleur [7K]2 years ago

6 0

Answer:

Step-by-step explanation:

The answer is no, because each value in the domain is associated with exactly one point in the codomain, but a line passing through the circle generally intersects the circle at two points.

This sort of thing is inconvenient, because circles are very important in geometry. Sometimes the points of a circle are described by a relation, given by (x−a)2+(y−b)2=r2, where (a,b) is the center and r is the radius. Because of the squares, there can be two different values of y that make the relation true for various values of x, so the graph of the relation is a circle. I will not suggest taking my answer for i have just started learning functions

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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:

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

5 0

2 years ago

Sheri bought flowers for centerpieces for a party. Which estimation strategies could she use to find the gratuity for the floris

Ivan

Answer: round the cost 90 and percent to 10 use a benchmark percent 0f to 10

Step-by-step explanation: took the test

4 0

2 years ago

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3. At the beginning of the day, a water tank contained 526.8 gallons of water. During the day,

Andru [333]

Answer:

208.85

Step-by-step explanation:

It is 208.85 because when you have .8 + .05 =.85

5 0

2 years ago

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Help ASAP ! I’ll mark as brainlist

mixas84 [53]

Answer:

1031 Meters

Step-by-step explanation:

You would use the Pythagorean Theorem to solve it which would be a^2 + b^2 = c^2 and a would be 800 and b would be 650 in this circumstance you would try to find C which would be glenn blvd.

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2 years ago

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According to the rational root theorem, what are all the potential rational roots of f(x)= 5x^3-7x+11

Tems11 [23]

The rational root theorem states that the rational roots of a polynomial can only be in the form p/q, where p divides the constant term, and q divides the leading term.

In your case, both the leading term 5 and the constant term 11 are primes, so their only divisors are 1 and themselves.

So, the only feasible solutions are

$\pm\dfrac{11}{5},\quad\pm 11,\quad \pm\dfrac{1}{5}, \pm 1$

For the record, in this case, none of the feasible solutions are actually a root of the polynomial.

8 0

2 years ago