The statements true about the the function f(x) = 2x2 – x – 6 are-
- The vertex of the function is (one-quarter, negative 6 and one-eighth).
- The function has two x-intercepts.
<h3>What is vertex of parabola?</h3>
The vertex of parabola is the point at the intersection of parabola and its line of symmetry.
Now the given function is,
f(x) = 2x^2 – x – 6
Also, it is given that the vertex is located at (0.25, -6) and the parabola opens up, the function has two x-intercepts.
Comparing the given function with standard form,
f(x) = a x^2 bx + c
By comprison we get,
a = 2
b = -1
c = -6
Now, x-coordinate of vertex is given as,
x = -b/2a
put the values we get,
x = -(-1)/2*2
or, x = 1/4
Put the value of x in given function, so y-coordinate of the vertex is given as,
f(1/4) = 2(1/4)² - 1/4 - 6
= -49/6
= -6 1/8
Hence, The statements true about the the function f(x) = 2x2 – x – 6 are-
- The vertex of the function is (one-quarter, negative 6 and one-eighth).
- The function has two x-intercepts.
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X+y=155. Y= -x+155
1.50x +4(-x+155)= 520
1.50x-4x+620=520
-2.5x =-100
X= 40. Y=110
Answer:
<em>A.</em>
<em>The student made an error in step 3 because a is positive in Quadrant IV; therefore, </em>
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Step-by-step explanation:
Given
Required
Where and which error did the student make
Given that the angle is in the 4th quadrant;
The value of r is positive, a is positive but b is negative;
Hence;
Since a belongs to the x axis and b belongs to the y axis;
is calculated as thus
Substitute
Rationalize the denominator
So, from the list of given options;
<em>The student's mistake is that a is positive in quadrant iv and his error is in step 3</em>
Answer:
Step-by-step explanation:
Component form of a vector is given by 
, where 
represents change in x-value and 
represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector 
, the magnitude is 
.
190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being , one leg being , and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.
In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.
Therefore, we have:
To find the other leg, , we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:
Verify that 
Therefore, the component form of this vector is 
