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ser-zykov [4K]
3 years ago
6

600,000+80,000+10 in a form

Mathematics
2 answers:
Nezavi [6.7K]3 years ago
7 0
680,010 is the answer
Iteru [2.4K]3 years ago
5 0
Standard is 680,010.Word is six hundred thousand eighty thousand and ten
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4. Diego is thinking of two positive numbers. He says, "If we triple the first number and
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Answer:

3<em>x </em>+ 2<em>y</em> = 34. and two possible pairs of positive numbers are  (<em>x</em>, <em>y</em>) = (10, 2) and (<em>x</em>, <em>y</em>) = (4, 11).

Step-by-step explanation:

 Let the First positive number be <em>x</em> and second positive number be <em>y.</em>

Triple of first number = 3<em>x</em>

double of second number = 2<em>y</em>

According to question,

3<em>x </em>+ 2<em>y</em> = 34

Therefore, the equation is 3<em>x </em>+ 2<em>y</em> = 34

So the two possible pair of numbers Diego would be thinking of must satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34

Now,  3<em>x </em>+ 2<em>y</em> = 34

3<em>x </em>= 34 - 2<em>y</em>

Let x = 10 and by substituting its value in above expression,

3 \times 10 = 34 - 2y

30 = 34 - 2y

2y = 34 - 30

2y = 4

y = 2

Therefore first pair (<em>x</em>, <em>y</em>) = (10, 2)

In the same way put x = 4 then,

3<em>x </em>= 34 - 2<em>y</em>

3 \times 4 = 34 - 2y

2y = 34 - 12

2y = 22

y = 11

Therefore first pair (<em>x</em>, <em>y</em>) = (4, 11)

Therefore, (<em>x</em>, <em>y</em>) = (4, 11) and  (<em>x</em>, <em>y</em>) = (10, 2) are the two possible pairs of numbers Diego could be thinking of as these both values satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34.  

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2 years ago
What is the value of b2 - 4ac for the following equation? 2x 2 - 2x - 1 = 0
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