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kow [346]
2 years ago
6

Classify the following as a fraction, expression, equation, or inequality b - 7 < 17

Mathematics
1 answer:
insens350 [35]2 years ago
3 0

D. Inequality

An expression would just contain simple arithmetic without any symbols relating two things to another.

An equation must contain an equals sign.

A fraction will have a bar representing division.

An inequality contains a sign representing a comparison of magnitude, in this question less than <.

Hope this helps!!

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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 the volume of a cube is 216cm fond the length of the cube ​
deff fn [24]

Answer:

6cm

Step-by-step explanation:

7 0
3 years ago
The library is 1.75 miles directly north of the school. The park is 0.6 miles directly south of the school. How far is the libra
kumpel [21]
If something is placed the way these places are(directly NORTH and the park is SOUTH) all you have to do is add your two values, lining up your decimal points. 1.75
            +.6
            2.35
Let me know if this helps by making me your brainliest answer
4 0
3 years ago
Read 2 more answers
4 cos²x – 1=0<br><br> What is the solution?
andreev551 [17]

Answer:

$x=\frac{\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{5\pi }{3}+2\pi n, n \in \mathbb{Z}$

$x=\frac{2\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{4\pi }{3}+2\pi n, n \in \mathbb{Z}$

or

$x=\frac{\pi}{3}+\pi n, n \in \mathbb{Z} $

$x=\frac{2\pi }{3}+\pi n, n\in \mathbb{Z}$

Step-by-step explanation:

4\text{cos}^2(x)-1=0\\4\text{cos}^2(x)=1\\

$cos(x)=\pm\sqrt{\frac{1}{4} } $

$cos(x)=\pm\frac{1}{2}  $

So, when cos(x) is equal to

$\frac{1}{2}   \text{ and } -\frac{1}{2}$    ?

For

$cos(x)=\frac{1}{2} $

We are talking about x = 60º and x = 300º, Quadrant I and IV, respectively. In radians:

$x=\frac{\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{5\pi }{3}+2\pi n, n \in \mathbb{Z}$

or

$x=\frac{\pi}{3}+\pi n, n \in \mathbb{Z} $

For

$cos(x)=-\frac{1}{2} $

We are talking about x = 120º and x = 240º, Quadrant II and III, respectively. In radians:

$x=\frac{2\pi }{3}+2\pi n, n\in \mathbb{Z}$

$\:x=\frac{4\pi }{3}+2\pi n, n \in \mathbb{Z}$

or

$x=\frac{2\pi }{3}+\pi n, n\in \mathbb{Z}$

6 0
3 years ago
Solve the following quadratic equation. Express your answers in exact form.
wlad13 [49]

Answer:

x = (2+√10)/ 3 and x= (2-√10/3)

Step-by-step explanation:

g(x) = -3x² +4x +2

use the quadratic formula x= (-b±√b²-4ac)/2a, when g(x) = ax² +bx +c

x= [-4 ±√4²-4*(-3)*2] / 2*(-3) , solve under the square root

x= (-4 ± √16 +24) / -6 , add under the radical

x= (-4 ±√40) / -6, simplify the square root

x= (-4 ±2√10) /-6, factor -2

x= -2 (2±√10)/-2 *3, simplify -2

x= (2±√10)/3

Therefore,

x = (2+√10)/ 3 and x= (2-√10/3)

5 0
3 years ago
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