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Fantom [35]
3 years ago
8

Elizabet is selling ham at a farmers market

Mathematics
1 answer:
Alinara [238K]3 years ago
5 0
Vedunejdjwndhje


AND THEN WHAT
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Can someone help me!!
4vir4ik [10]
I think ur supposed to r
4 0
2 years ago
The graph of F(x) shown below resembles the graph of G(x) = x ^ 2 but it has been changed somewhat . Which of the following coul
sukhopar [10]

Answer:

f(x) = 3x^{2} + 2

Step-by-step explanation:

Given G(x) = x^{2}

As we can see, the graph of f(x) is 2 units above the graphof g(x) in vertical (vertical shift)

=> f(x) = ax^{2} + 2

As you can see, the graph of f(x) is stretched vertically by a dilation factor of 3

and there is no reflection, so a= 3

=> the  equation of f(x) is:  f(x) = 3x^{2} + 2

5 0
3 years ago
Find the equation in standard form of the line passing through the points 3,-4 and 5,1
aliina [53]

Answer:

5x - 2y - 23 = 0

Step-by-step explanation:

Line is passing through the points (3,\:-4)= (x_1, \:y_1) \: \&\: (5,\: 1)= (x_2,\:y_2)

Equation of line in two point form is given as:

\frac{y -y_1 }{y_1 -y_2 }  =  \frac{x -x_1 }{x_1 -x_2 } \\  \\  \therefore \:  \frac{y -( - 4) }{ - 4 -1}  =  \frac{x -3 }{3 -5 } \\  \\  \therefore \:\frac{y  +  4 }{ - 5}  =  \frac{x -3 }{ - 2 } \\  \\  \therefore \: \frac{y  +  4 }{  5}  =  \frac{x -3 }{  2 } \\  \\  \therefore \:  2(y + 4) = 5(x - 3) \\ \therefore \:  2y + 8 = 5x - 15 \\ \therefore \: 5x - 15 - 2y - 8 = 0 \\  \red{ \boxed{ \bold{\therefore \: 5x - 2y - 23 = 0}}} \\ is \: the \: required \: equation \: of \: line \: in \:  \\ standard \: form.

7 0
3 years ago
Wich symbol makes the following statement true ?
azamat
The answer is < because 431,511 is more than 413,115

4 0
4 years ago
Read 2 more answers
usa el teorema de la altura para proponer como se podria construir un segmento cuya longitud sea media proporcional entre dos se
Mrac [35]
Usando el teorema de altura 

El teorema de altura relaciona la altura (h) de un triángulo rectángulo (ver figura) y los catetos de dos triángulos que son semejantes al anterior ABC, al trazar la altura (h) sobre la hipotenusa. De manera que e<span>n todo </span>triángulo rectángulo, la altura (h<span>) relativa a la </span>hipotenusa<span> es la </span>media geométrica<span> de las dos proyecciones de los </span>catetos<span> sobre la </span>hipotenusa<span> (</span>n<span> y </span>m<span>). Es decir, se cumple que:

</span>\frac{n}{h}= \frac{h}{m}

Dado que el problema establece <span>construir un segmento cuya longitud sea media proporcional entre dos segmentos de 4 y 9 cm, entonces, digamos que n = 4cm y m = 9cm tenmos que:

</span>\frac{4}{h}= \frac{h}{9}

De donde:

h= \sqrt{(4)(9)}=\sqrt{36}=6cm

¿Cómo se podria construir si los segmentos son de a cm y b cm?

Si los segmentos son de a y b cm entonces a y b son parámetros que pueden tomar cualquier valor positivo siempre que se cumpla que:

h= \sqrt{ab}

6 0
3 years ago
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