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otez555 [7]
3 years ago
9

Mo and Jo Share Their "Dough" Mo and Jo each have a different amount of money. They only have dollar bills. If Mo gave Jo $10, t

hey would both have the same amount. If, instead, Jo gave Mo $10, Mo would have twice as much money as Jo. HOW MUCH MONEY DO THEY EACH HAVE AND HOW MUCH DO MO AND JO HAVE ALTOGETHER?
help please :)
Mathematics
1 answer:
Zinaida [17]3 years ago
8 0

Answer:

Mo has $70, Jo has $50 and together they have $120.

Step-by-step explanation:

As the statement indicates that if Mo gave Jo $10, they would both have the same amount, you can say that you can subtract $10 from the money Mo has and that would be equal to the amount Jo has plus $10, which can be expressed as:

x-10=y+10 (1), where:

x is the amount of money Mo has

y is the amount of money Jo has

Also, the statement says that if Jo gave Mo $10, Mo would have twice as much money as Jo which means that if you add 10 to the money Mo has this would be equal to two times the money Jo has minus $10, which would be:

x+10=2(y-10) (2)

Now, you can isolate x in the first equation and then, replace that on the second one:

x=y+20

Now, you can replace this in 2 and solve for y:

x+10=2(y-10)

y+20+10=2y-20

30+20=2y-y

y=50

You can replace the value of y in x=y+20 in order to find the value of x:

x=50+20

x=70

Finally, you can add both amounts to find the amount of money they have altogether:

70+50=120

According to this, the answer is that: Mo has $70, Jo has $50 and together they have $120.

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If d , e, and f are midpoints of the sides of ABC, find the perimeter of ABC.
ehidna [41]

Answer:

Perimeter of the ΔDEF = 10.6 cm

Step-by-step explanation:

The given question is incomplete; here is the complete question with attachment enclosed with the answer.

D, E, and F are the midpoints of the sides AB, BC, and CA respectively. If AB = 8 cm, BC = 7.2 cm and AC = 6 cm, then find the perimeter of ΔDEF.

By the midpoint theorem of the triangle,

Since D, E, F are the midpoints of the sides AB, BC and CA respectively.

Therefore, DF ║ BC and FD=\frac{1}{2}\times(BC)

FD = \frac{7.2}{2}

     = 3.6

Similarly, FE=\frac{1}{2}\times(AB)

FE=\frac{8}{2}

FE = 4 cm

And DE=\frac{AC}{2}

DE = \frac{6}{2}

     = 3 cm

Now perimeter of ΔDEF = DE + EF + FD

= 3 + 4+ 3.6

= 10.6 cm

Perimeter of the ΔDEF is 10.6 cm.

7 0
3 years ago
Al sumar y/o restar números decimales, siempre se debe aplicar la norma de los signos. * 1 punto FALSO VERDADERO
Luden [163]

Answer:

La norma de los signos es para el producto de números reales, y la norma es la siguiente.

(+)*(+) = (+)

(+)*(-) = (-)

(-)*(+) = (-)

(-)*(-) = (+)

Es decir, el producto de dos números de mismo signo es siempre positivo

El producto de dos números de distinto signo es siempre negativo.

Particularmente, para la suma esta norma no funciona (pues no está definida para la suma)

Pero en casos como:

5 - (-4)  

(esto sería: "la diferencia entre cinco y menos cuatro")

notar que podemos reescribir esto como:

5 + (-1)*(-4)

Ahora podemos aplicar la norma de los signos:

5 + 4 = 9

Donde aplicamos la norma de los signos,  

Podemos concluir que, si bien es una regla que aplica al producto, siempre la tenemos que tener en cuenta en cualquier operación que hagamos.

Por lo podemos concluir que la respuesta correcta es verdadero.

7 0
3 years ago
Zaid has a peculiar pair of four-sided dice. When he rolls the dice, the probability of any particular outcome is proportional t
Elden [556K]

Answer:   a) \bold{\dfrac{3}{16}}     b) \bold{\dfrac{1}{36}}

<u>Step-by-step explanation:</u>

a) In order to get an even number, you have the 3 different scenarios:

1) Even, Even, Even, Even     \dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}

2) Even, Even, Odd, Odd   \dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}

3) Odd, Odd, Odd, Odd   \dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}

<em>Order doesn't matter</em>

Add them up to get your answer: \dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\quad =\large\boxed{\dfrac{3}{16}}

b) If one die is a 2 and another is a 3 and the other two dice can be any number, then you have 1 possibility for a 2, 1 possibility for a 3, and 6 possibilities for each of the other two dice.

\dfrac{1\times 1\times 6\times 6}{6^4}\quad =\dfrac{1}{6^2}\quad =\large\boxed{\dfrac{1}{36}}

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What is 1 2/3 as an improper fraction
Troyanec [42]
It is 5/3 ......................
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Read 2 more answers
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vagabundo [1.1K]

Answer:

5.49

Step-by-step explanation:

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3 0
3 years ago
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