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3 years ago

12

An insecticible contains 85 centigrams of inert ingredient for every 1 centigram of active ingredient. if a quantity of the weig

hs 688 centigrams, how much of each type of ingredient does it contain?

1 answer:

11111nata11111 [884]3 years ago

7 0

(85/86)*688 cg = 680 cg . . . inert ingredient
.. 688 cg -680 cg = 8 cg . . . .active ingredient

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There are 9 girls and 11 boys in a class at a Private

blagie [28]

Answer:

35%

Step-by-step explanation:

9+11=20

3+4=7

20*5=100

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3 years ago

CUALES SON LOS GRADOS PARA HACER UNA GRAFICA DE PASTEL, CON LOS SIGUIENTES PORCENTAJES( 20%,50%,10%)

svlad2 [7]

Answer:

Lo primero que debemos saber, es que la gráfica de pastel nos permite dividir a una población, la cual sería el 100%, en porcentajes menores que representan algo.

En este caso tenemos los porcentajes 20%, 50%, 10% (notar que no suman 100%, entonces nos falta algún porcentaje)

Bueno, como sabemos el gráfico (el cual es un círculo) representa el 100%.

Y en un círculo, tenemos un ángulo de 360°

Entonces comenzamos con la equivalencia:

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Entonces, ¿a que grado corresponde el 50%? Planteamos la ecuación:

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obtenemos:

x/360° = 50%/100%

resolviendo para x, obtenemos:

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Para los otros porcentajes hacemos lo mismo.

Para el de 20% tenemos:

y = 20%

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Finalmente, para el de 10% tendremos:

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3 years ago

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Nataly_w [17]

Answer

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5 0

2 years ago

Read 2 more answers

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Elina [12.6K]

Answer:

Look that the coordinates and trace those points to a corresponding number on the x or y axis. Then write the first number in the coordinate in the x value and the second number for the y value, repeat the process for the rest of the points.

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Please help! This assignment is due soon! (In The photo)

eimsori [14]

Answer

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Then Elena must have substracted 1/2x from both sides of the equation.

That is;

$\begin{gathered} \frac{1}{2}x+3=\frac{7}{2}x+5 \\ \text{Substracting }\frac{1}{2}x\text{ from both sides of the equation will give Elena first step} \\ \frac{1}{2}x+3-\frac{1}{2}x=\frac{7}{2}x+5-\frac{1}{2}x \\ 3=\frac{7}{2}x-\frac{1}{2}x+5 \end{gathered}$

For Lin to have arrived at

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It shows Lin must have multiplied both sides of the equation by 2

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$\begin{gathered} \frac{1}{2}x+3=\frac{7}{2}x+5 \\ \text{Multiply both sides of the equation by the lowest common mutiple } \\ \text{of the denominator which is 2.} \\ \frac{1}{2}x(2)+3(2)=\frac{7}{2}x(2)+5(2) \\ x+6=7x+10 \end{gathered}$

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1 year ago