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wolverine [178]
4 years ago
13

sara quiere guardar 48 manzanas y 36 duraznos en cajas con el mismo numero de frutas de cada tipo en cada caja cuantas cuantas f

rutas pude guardar teniendio en cuenta que nesecita el mayor numero de cajas posibles
Mathematics
1 answer:
coldgirl [10]4 years ago
7 0

Answer:

Se requiere 12 cajas con 4 manzanas y 3 duraznos cada una.

Step-by-step explanation:

Se requiere el menor número posible de manzanas y duraznos para obtener el mayor número de cajas. Puesto que no se considera cortar fruta alguna en porciones y sabiendo que hay más manzanas que duraznos, se determina la razón mínima de manzanas por duraznos:

x = \frac{48\,manzanas}{36\,duraznos}

x = \frac{4 manzanas}{3\,duraznos}

Se requiere 4 manzanas por cada 3 duraznos. Entonces, cada caja debe contener 3 duraznos y 4 manzanas. El número máximo de cajas es:

n = \frac{48\,manzanas}{4\,\frac{manzanas}{caja} } = \frac{36\,duraznos}{3\,\frac{duraznos}{caja} }

n = 12\,cajas

Se requiere 12 cajas con 4 manzanas y 3 duraznos cada una.

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Oduvanchick [21]

Answer:

20

Step-by-step explanation:

(90+70)-(80+60)=20

just minus the total selling price to the total cost price

3 0
3 years ago
14 divied by 3/8 equals
astraxan [27]

Answer:

37 1/3

Step-by-step explanation:

14 divided by 3/8=

14*8/3=

37 1/3

5 0
3 years ago
Multiply and simplify the following complex numbers? (4- 2i).(-5 +4i)
Ronch [10]

Answer:

-12 +26i

Step-by-step explanation:

Multiple using the FOIL Method, then combine the real and imaginary parts of the expression.

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7 0
3 years ago
For how many positive values of n are both n3 and 3n four-digit integers?
AVprozaik [17]

Answer:

No positive value of n

Step-by-step explanation:

we have to find out for how many positive values of n are both

n^3 , 3^n our-digit integers

Let us consider first cube

we get 4digit lowest number is 1000 and it has cube root as 10.

Thus 10 is the least integer which satisfies four digits for cube.

The highest integer is 9999 and it has cube root as 21.54

or 21 the highest integer.

Considering 3^n we get,

3^10 is having 5 digits and also 3^21

Thus there is no positive value of n which satisfy that both n cube and 3 power n are four digits.

5 0
4 years ago
Solve using the quadratic formula<br> 2x^2+x+67=0
telo118 [61]

Answer:

\displaystyle x=\frac{-1 \pm i\sqrt{535}}{2}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Multiple Roots
  • Standard Form: ax² + bx + c = 0
  • Quadratic Formula: \displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

<u>Algebra II</u>

  • Imaginary Root <em>i</em> = √-1

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

2x² + x + 67 = 0

<em>a</em> = 2

<em>b</em> = 1

<em>c</em> = 67

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Substitute in variables [Quadratic Formula]:                                                  \displaystyle x=\frac{-1 \pm \sqrt{1^2-4(2)(67)}}{2(1)}
  2. Multiply:                                                                                                             \displaystyle x=\frac{-1 \pm \sqrt{1^2-4(2)(67)}}{2}
  3. [√Radical] Evaluate exponents:                                                                       \displaystyle x=\frac{-1 \pm \sqrt{1-4(2)(67)}}{2}
  4. [√Radical] Multiply:                                                                                           \displaystyle x=\frac{-1 \pm \sqrt{1-536}}{2}
  5. [√Radical] Subtract:                                                                                          \displaystyle x=\frac{-1 \pm \sqrt{-535}}{2}
  6. [√Radical] Simplify:                                                                                           \displaystyle x=\frac{-1 \pm i\sqrt{535}}{2}
3 0
3 years ago
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