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3 years ago

5

A pound of turkey costs (3w 8) dollars and a pound of cheese costs (4w-3) dollars.mrs.young bought 2 pounds of turkey and three

pounds of cheese.

1 answer:

zlopas [31]3 years ago

3 0

$x = 2(3w + 8) + 3(4w - 3)$ , where x is the total cost in terms of w.

$x = (6w + 16) + (12w - 9)$

$x = 18w + 7$

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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

Leno4ka [110]

H = hrs she worked
p = phone calls she answered

h x 10 + p x .25 = maggies earnings

Part A
h x 10 + 60 x .25 = 115

Part B
h x 10 + 15 = 115 (multiplied 60 x .25)
h x10 = 100 (subtracted 15 from each side of the equation)
h = 10 (divided 10 into each side)

Part C
10 hrs

Hope this helps

3 0

3 years ago

Read 2 more answers

Please help ASAP<br>Will be given brainliest!!!!

zmey [24]

Answer:

C

Step-by-step explanation:

We need common denominator, which in this case is 40.

6*5 is 30 and 4*8 is 32.

We add these up 62/40

Now add the wholes, 5+4 is 9

We have an improper fraction, which will add 1 to 9, causing the answer to be C.

3 0

2 years ago

Considering only the values of β for which sinβtanβsecβcotβ is defined, which of the following expressions is equivalent to sinβ

-Dominant- [34]

Answer:

$\tan(\beta)$

Step-by-step explanation:

For many of these identities, it is helpful to convert everything to sine and cosine, see what cancels, and then work to build out to something. If you have options that you're building toward, aim toward one of them.

${\tan(\theta)}={\dfrac{\sin(\theta)}{\cos(\theta)}$ and ${\sec(\theta)}={\dfrac{1}{\cos(\theta)}$

Recall the following reciprocal identity:

$\cot(\theta)=\dfrac{1}{\tan(\theta)}=\dfrac{1}{ \left ( \dfrac{\sin(\theta)}{\cos(\theta)} \right )} =\dfrac{\cos(\theta)}{\sin(\theta)}$

So, the original expression can be written in terms of only sines and cosines:

$\sin(\beta)\tan(\beta)\sec(\beta)\cot(\beta)$

$\sin(\beta) * \dfrac{\sin(\beta) }{\cos(\beta) } * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) } {\sin(\beta) }$

$\sin(\beta) * \dfrac{\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} {\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}$

$\sin(\beta) *\dfrac{1 }{\cos(\beta) }$

$\dfrac{\sin(\beta)}{\cos(\beta) }$

Working toward one of the answers provided, this is the tangent function.

The one caveat is that the original expression also was undefined for values of beta that caused the sine function to be zero, whereas this simplified function is only undefined for values of beta where the cosine is equal to zero. However, the questions states that we are only considering values for which the original expression is defined, so, excluding those values of beta, the original expression is equivalent to $\tan(\beta)$ .

8 0

2 years ago

Please help.<br> Algebra.

Akimi4 [234]

1. A

Since parallel lines never cross, then there can be no intersection; that is, for a system of equations that graphs as parallel lines, there can be no solution. This is called an "inconsistent" system of equations, and it has no solution.

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2. C

That's right. If a system of equations has a solution, then their graphs intersect, and the point where they intersect is the solution because it's the point that satisfies each equation in the system.

Straight-line graphs with the same slope are parallel lines, and they never intersect, which is another way of saying they have no solution.

8 0

3 years ago