Answers:
- C) x = plus/minus 11
- B) No real solutions
- C) Two solutions
- A) One solution
- The value <u> 18 </u> goes in the first blank. The value <u> 17 </u> goes in the second blank.
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Explanations:
- Note how (11)^2 = (11)*(11) = 121 and also (-11)^2 = (-11)*(-11) = 121. The two negatives multiply to a positive. So that's why the solution is x = plus/minus 11. The plus minus breaks down into the two equations x = 11 or x = -11.
- There are no real solutions here because the left hand side can never be negative, no matter what real number you pick for x. As mentioned in problem 1, squaring -11 leads to a positive number 121. The same idea applies here as well.
- The two solutions are x = 0 and x = -2. We set each factor equal to zero through the zero product property. Then we solve each equation for x. The x+2 = 0 leads to x = -2.
- We use the zero product property here as well. We have a repeated factor, so we're only solving one equation and that is x-3 = 0 which leads to x = 3. The only root is x = 3.
- Apply the FOIL rule on (x+1)(x+17) to end up with x^2+17x+1x+17 which simplifies fully to x^2+18x+17. The middle x coefficient is 18, while the constant term is 17.
Answer:
120 sq. units
plus them all and add 100
Answer:
Are you 1 grade or what......
There is no hard and fast rule to select the class width. It largely depends on our application.However, one thing that should be kept in mind is that the number of classes should neither to be too low nor too high. So keeping this thing in mind, the class width is select.
The range of the data is = Maximum- Minimum = 96 - 11 = 85
10 classes will be most suited for this data.
The class width for each data can be calculated as:
Class Width = Range / Number of Classes = 85/10 = 8.5
Class width is always rounded to nearest next integer. So the class width will be 9 in this case.
So, the best value of class width or interval width for the given data will be 9.
Answer:
- $8000 at 1%
- $2000 at 10%
Step-by-step explanation:
It often works well to let a variable represent the amount invested at the higher rate. Then an equation can be written relating amounts invested to the total interest earned.
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<h3>setup</h3>
Let x represent the amount invested at 10%. Then 10000-x is the amount invested at 1%. The total interest earned is ...
0.10x +0.01(10000 -x) = 280
<h3>solution</h3>
Simplifying gives ...
0.09x +100 = 280
0.09x = 180 . . . . . . . subtract 100
x = 2000 . . . . . . divide by 0.09
10000 -x = 8000 . . . . amount invested at 1%
<h3>1.</h3>
$8000 should be invested in the 1% account
<h3>2.</h3>
$2000 should be invested in the 10% account
