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zimovet [89]
2 years ago
9

Help me .............​

Mathematics
1 answer:
Vlad [161]2 years ago
4 0

Answer:#1 = 16

#2 = 4

Step-by-step explanation:

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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
zavuch27 [327]

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
If 2x+20 and 3x+20 makes a straight angke find value of x​
erma4kov [3.2K]

Answer:

x = 28

Step-by-step explanation:

The sum of the angles in a straight angle = 180°

Sum the angles and equate to 180

2x + 20 + 3x + 20 = 180

5x + 40 = 180 ( subtract 40 from both sides )

5x = 140 ( divide both sides by 5 )

x = 28

3 0
3 years ago
Julia has 2 identical rooms in her house. if each room measures 8 feet on one side and 12 feet on another, what is the total are
Jlenok [28]
First we have the work out the area of 1 room. So that is 8ft * 12ft which is 96ft^2. And if there are 2 of these rooms, then we multiply this value by 2. Giving us a total area of 192ft^2.
5 0
3 years ago
If f(x) = x2 + 4 and g(x) = <img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D" id="TexFormula1" title="\sqrt{x}" alt="\sqrt{x}" a
dangina [55]

Answer: 12

Step-by-step explanation:

6 0
2 years ago
A plane flying horizontally at an altitude of 1 mi and a speed of 470 mi/h passes directly over a radar station. Find the rate a
kiruha [24]

Answer:

407 mi/h

Step-by-step explanation:

Given:

Speed of plane (s) = 470 mi/h

Height of plane above radar station (h) = 1 mi

Let the distance of plane from station be 'D' at any time 't' and let 'x' be the horizontal distance traveled in time 't' by the plane.

Consider a right angled triangle representing the above scenario.

We can see that, the height 'h' remains fixed as the plane is flying horizontally.

Speed of the plane is nothing but the rate of change of horizontal distance of plane. So, s=\frac{dx}{dt}=470\ mi/h

Now, applying Pythagoras theorem to the triangle, we have:

D^2=h^2+x^2\\\\D^2=1+x^2

Differentiating with respect to time 't', we get:

2D\frac{dD}{dt}=0+2x\frac{dx}{dt}\\\\\frac{dD}{dt}=\frac{x}{D}(s)

Now, when the plane is 2 miles away from radar station, then D = 2 mi

Also, the horizontal distance traveled can be calculated using the value of 'D' in equation (1). This gives,

2^2=1+x^2\\\\x^2=4-1\\\\x=\sqrt3=1.732\ mi

Now, plug in all the given values and solve for \frac{dD}{dt}. This gives,

\frac{dD}{dt}=\frac{1.732\times 470}{2}=407.02\approx 407\ mi/h

Therefore, the distance from the plane to the station is increasing at a rate of 407 mi/h.

6 0
3 years ago
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