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2 years ago

con 16 barras de mantequilla se prepararon 20 postres ?que fraccion del total de mantequilla corresponde a cada postre

Mathematics

1 answer:

MatroZZZ [7]2 years ago

5 0

De ahí que la fracción de la mantequilla total que corresponde a cada postre sea 1/4

Si con 16 barras de mantequilla se prepararon 20 postres, esto se puede expresar como:

16 pegatinas = 20 postres

Para obtener la fracción del total de mantequilla que corresponde a cada postre, podemos escribir;

x = 1 postre

Divide ambas expresiones

16 / x = 20/1

20 veces = 16 * 1

20x = 16

x = 16/20

x = 4/5

De ahí que la fracción de la mantequilla total que corresponde a cada postre sea 1/4

Obtenga más información aquí: brainly.com/question/1878884

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Answer:

a) $P=3(120-s)+200$

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Step-by-step explanation:

For this case we can use a linear model to solve the problem.

s) Create an equation to express the increase on the price tickets and the number of seats sold

$s$ number of seats, if w analyze the info given the number of seats after increase the price is given by $120-s$ .

And let P the price for the ticket. So after the increase in ticket price the expression for the increase is P-200.

We have an additional info, for each increase of $3 the number of setas decrease 1. And the equation that gives to us the price change in terms of the increase of price is:

$P-200=3(120-s)$

So then our linear equation is given by:

$P=3(120-s)+200$

b) Over a certain period, the number of seats sold for this flight ranged between 90 and 115. What was the corresponding range of ticket prices?

So for this case we just need to replace the limits into the linear equation and see what we got:

$P_L=3(120-90)+200=290$

$P_U=3(120-115)+200=215$

So the corresponding range of ticket prices is:

$215 \leq P \leq 290$

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3 years ago

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dangina [55]

$\boxed {y= \frac{2}{3} x - 17}$

$Slope \longrightarrow \frac{2}{3}$

$Y-intercept \longrightarrow -17$

$\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow$

$\boxed {4x-6y=-6}$

$Rearranging\ it \ to \ make \ y \ the \ subject$

$6y=4x +6$

$y= \frac{2}{3} x +1$

$Slope \longrightarrow \frac{2}{3}$

$Y-intercept \longrightarrow 1$

$\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow\downarrow$

$\boxed{Both \ the \ slope \ are \ the \ same \longrightarrow They \ are \ parallel }$

$\boxed {The \ y-intercepts \ are \ different

\longrightarrow They \ are \ not \ the \ same \ line }$

$\boxed{\boxed{Answer B}}$

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