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eduard
3 years ago
8

estaba previsto destinar 3/14 de partes de una finca a plazas de aparcamiento pero finalmente han destinado 3/4 de lo previsto a

zonas ajardinadas que fraccion de la finca se ha destinado a plazas de aparcamiento
Mathematics
1 answer:
JulsSmile [24]3 years ago
8 0

Queremos ver que fracción de la finca se ha destinado a plazas de aparcamiento.

La solución es:

La fracción de la finca que se destina a plaas de aparcamiento es 3/56

Sabemos que originalmente se iba a destinar 3/14 del total de la finca a plazas de aparcamiento, pero finalmente se destino 3/4 de lo previsto a zonas ajardinadas.

Es decir, <u>se destino 3/4 de los 3/14 del total de la finca</u> a zonas ajardinadas, entonces <u>el 1/4 restante se dedico a plazas de aparcamiento</u>, esto da:

(1/4)*3/14  = 3/56

La fracción de la finca que se destina a plaas de aparcamiento es 3/56

Sí quieres aprender más, puedes leer:

brainly.com/question/16649102

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bogdanovich [222]
Hello,

The answer is C, refer to this picture for an explanation.
Have a great day!!
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3 years ago
Which is equivalent to ?
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I do not see an illustration.

8 0
3 years ago
In ΔVWX, the measure of ∠X=90°, WX = 8.3 feet, and XV = 2.5 feet. Find the measure of ∠W to the nearest tenth of a degree.
kiruha [24]

Answer:

16.76°

Step-by-step explanation:

In ΔVWX, the measure of ∠X=90°, WX = 8.3 feet, and XV = 2.5 feet.

We want to find the measure of <W.

We know side length that is adjacent and opposite to <W.

We can use the tangent ratio, to find the measure of <W.

The tangent ratio is opposite over hypotenuse.

\tan(m \angle \: w)  =  \frac{2.5}{8.3}

\tan(m \angle \: w)  =  0.301

Take tangent inverse to get:

m \angle \: w= { \tan}^{ - 1}   (0.301)

m \angle \: w=16.76  \degree

7 0
3 years ago
A 9 kilogram bag of animal feed costs $45. ​ What is the unit price per kilogram?
kotykmax [81]
The unit price per kg is $5, to find this divide the 9 kilograms into the price $45 the answer is what each kilogram is worth
6 0
3 years ago
what is the solution to set to the inequality (4x-3)(2x-1)_&gt;0 a) {x|x&lt;_3 or x_&gt;1} b) {x|x&lt;_2 or x_&gt; 4/3} c) {x|x&
Vinil7 [7]
       (4x - 3)(2x - 1) ≥ 0
4x - 3 ≥ 0    or    2x - 1 ≤ 0
    + 3  + 3                + 1 + 1
     4x ≥ 3                 2x ≤ 1
      4     4                  2    2
       x ≥ ³/₄                  x ≤ ¹/₂
        x ∈ [-∞, ¹/₂] ∧ [³/₄, ∞]

The answer is C.
6 0
3 years ago
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