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skelet666 [1.2K]
3 years ago
9

Little John had $8.50. He spent $1.25 on sweets and gave to his two friends $1.20 each. How much money was left?

Mathematics
2 answers:
lesya692 [45]3 years ago
8 0

Answer:

<h2>$4.85</h2>

Step-by-step explanation:

John spent and gave to his two friends a total of

1.25 + 1.20 + 1.20 = $3.65

Money left

8.50 - 3.65 = $4.85

-------------------------------------------------------------------------------------------------------------

Thanks!

Mark me brainliest!

~FieryAnswererGT~

kaheart [24]3 years ago
6 0

Answer:

$4.85

Step-by-step explanation:

Given;

Little John had $8.50

Spent $1.25 on sweets

Gave 2 friends $1.20

Step 1; [Solve Spend on Sweets]

$8.50 - $1.25=7.25

Step 2; Find amount given to his 2 friend]

[Gave $1.20 to each of his 2 friends]

$1.20 x 2 = 2.40

7.25 - 2.40=4.85  

Hence, Little John have $4.85 is left

[RevyBreeze]

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cricket20 [7]

As soon as I read this, the words "law of cosines" popped
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you need for this problem.

The "law of cosines" relates the lengths of the sides of any
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since we know all the sides, and we want to find one of the angles.

To find angle-B, the law of cosines says

       b² = a² + c² - 2 a c cosine(B)

B  =  angle-B
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                  (1.4)² = (1)² + (1.9)² - (2 x 1 x 1.9) cos(B)

                 1.96  =  (1) + (3.61)  -  (3.8) cos(B)

Add  3.8 cos(B)  from each side:

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Subtract  1.96  from each side:

                             3.8 cos(B) =  2.65

Divide each side by  3.8 :

                                  cos(B)  =  0.69737  (rounded)

Whipping out the
trusty calculator:
                                 B  = the angle whose cosine is 0.69737

                                      =  45.784° .

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8 0
4 years ago
Me ayudan hacer este ejercicio con los pasos
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Answer:

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Step-by-step explanation:

8/7 plus 13/7. You add the numerators (the numbers on top) together. That makes 21. The denominator (the numbers on bottom) stays the same. So it would be 21/7. 7 fits into 21, 3 times.

8/7 más 13/7. Agrega los numeradores (los números en la parte superior) juntos. Eso hace 21. El denominador (los números en la parte inferior) permanece igual. Entonces sería 21/7. 7 encaja en 21, 3 veces.

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8/7 plus 13/7. Quarum numeratores addere (supra de numero) una. 21. Quod facit denominator est (per numeros in fundo) manebit. Ita esset 21/7. Vicium, in VII XXI, III tempora.

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