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Kay [80]
3 years ago
7

Estimate the sum to the nearest tenth: (-2.678) + 4.5 + (-0.68)

Mathematics
2 answers:
satela [25.4K]3 years ago
6 0
The best thing that would first be done in the aspect would be to first sum up the numbers that are in parenthesis.

Let's keep in mind that a (-)+(-) would equal a positive. This would mean that a negative number, plus another negative number would then be a positive number.

Your answer: <span>1.14200 </span>

prisoha [69]3 years ago
4 0
<span>(-2.678) + 4.5 + (-0.68)= 1.142
since .1 is in the tenth place and 42 isnt greater than 50, the answer would be 1.1. </span>
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Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
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Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term x^2, which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form f(x)=ax^2+bx+c is given by the expression: x_v=\frac{-b}{2a}

Since in our case a=1 and b=5, we get that the x-position of the vertex is: x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the x_v=-\frac{5}{2}. Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

5 0
3 years ago
Veronica wants to buy a house that costs $249,900. She is going to make a 20% down payment. How
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▹ Answer

<em>$49,980</em>

▹ Step-by-Step Explanation

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6 0
3 years ago
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<h3>What is the area of the trapezoid?</h3>

Generally, the equation for the trapezoid   is mathematically given as

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Step-by-step explanation:

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Divide:

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