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3 years ago

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Holland added three decimals that made a sum of 2. One of the decimals was 0.34. What are two other decimals Holland could have

used to make a sum of 2

2 answers:

bogdanovich [222]3 years ago

7 0

Using x to represent our missing values and the fact that there are two missing values gives us x + x = 2x. Because it's the sum of these numbers and 0.34, we get the equation 2x + 0.34; now we set it equal to 2, 2x + 0.34 = 2.

To solve, 1st subtract 0.34 on both sides to get 2x = 1.66. Now divide by 2 on both sides to get x = 0.83.

So your decimals are 0.34, 0.83, and 0.83

melamori03 [73]3 years ago

3 0

Answer:

In general the two other decimals are x and y that satisfy x + y = 1.66 where x ≤ .99 and y ≤ .99

One particular pair would be x = 0.99 and y = 0.67

Step-by-step explanation:

If Holland added three decimals that made a sum of 2 and one of them was 0.34, we are going to name x and y to the other 2 decimals.

We are going to write the equation:

$0.34 + x +y= 2\\x+y=2-0.34\\x+y = 1.66$

So we have that both the other numbers have to sum 1.66, however, we know both numbers are decimals so neither one of them can be greater than 1.

So this problem has many answers and the answers would be all the numbers that satisfy x + y = 1.66 where x ≤ .99 and y ≤ .99

For example, if x = 0.99, y would be:

y = 1.66 -0.99 = 0.67.

And our pair of numbers would be 0.99 and 0.67.

If x = .80, y would be

y = 1.66 - 0.80 = 0.86

And our pair of numbers would be 0.80 and 0.86

This way, we can find more than 1 answer for the other two decimals.

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ANSWER

$x=1.39$ to the nearest hundredth.

EXPLANATION

We have the equation,

$0=2x^2+3x-8$

Which can be rewritten as

$2x^2+3x-8=0$

The question demands the use of the quadratic formula.

Comparing this to the general quadratic equation;

$ax^2+bx+c=0$

$a=2,b=3,c=-8$

The quadratic formula is given by;

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

We now substitute all the values in to the formula;

$x=\frac{-3\pm\sqrt{(3)^2-4(2)(-8)}}{2(2)}$

We simplify to obtain;

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$x=\frac{-3\pm\sqrt{73}}{4}$

We now split the plus or minus sign to obtain;

$x=\frac{-3-\sqrt{73}}{4}$

This implies that;

$x=-2.89$

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$x=\frac{-3+\sqrt{73}}{4}$

This will evaluate to;

$x=1.39$

Hence, the positive solution rounded to the nearest hundredth is

$x \approx 1.39$

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