Incorrect, she should have divided both sides by 8. The answer should be x=7. There is only 1 solution to the equation because x is raised to the first degree. It is not a polynomial equation so there won't be more than 1 answer to the equation.
Answer:
a = 0.25
Step-by-step explanation:
Our strategy to solve this problem will be to use the information given in the table to obtain first the value of c in the quadratic equation which has the form ax^2 + bx + c and then form a system of 2 linear equations and solve for the coefficients a and b as follow:
from x=0 and y=-3 a*0 + b* 0 + c = - 3
c = -3
from x=1 and y= -3.75
a*(1) + *(1) + (-3) = -3.75
a + b = -3.75 + 3 = -0.75
from x= 2 and y = -4
a* (2)^2 + b*(2) + (-3) = -4
4a + 2b = -4 +1
4a + 2b = -1
Now we can solve the system f equations by elimination:
a + b = -0.75
4a + 2b = -1
multiply first equation by -2 and add to the second and get
-2a - 2b = +1.50
4a + 2b = -1
2a = 0.50 and substituting into any of the equations get b = -1
so the quadratic equation has a= 0.25 b= -1 c= -3
we can even plug the any of the other values for x given in the table and check the answer.
Answer:
y = -1/2 x
Step-by-step explanation:
Follow the directions "Complete the steps to write the equation of direct variation. Start with the equation of direct variation y = kx. Substitute in the given values for x and y to get . Solve for k to get . Write the direct variation equation with the value found for k."
y = kx substitute y = -4 and x = 8.
-4 = k*8
-4/8 = k
-1/2 = k
So the equation is y = -1/2(x).
Answer:

Step-by-step explanation:
<u>Theorem 1:</u> The height of right triangle drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of these two segments. Hence,

<u>Theorem 2:</u> In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.
Thus,
