Solve this equation d=2r solve for r

1 answer:

6 0

Let's solve for r.

d=2rStep 1: Flip the equation.

2r=dStep 2: Divide both sides by 2.

2r/2=d/2

r=1/2d

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What are the zeros of the quadratic function f(x) = 6x^2 + 12x – 7?

strojnjashka [21]

Answer:

$x1=\frac{-2+\sqrt{26/3}}{2}$

$x2=\frac{-2-\sqrt{26/3} }{2}$

Step-by-step explanation:

To find the zeros of the quadratic function f(x)=6x^2 + 12x – 7 we need to factorize the polynomial.

To do so, we need to use the quadratic formula, which states that the solution to any equation of the form ax^2 + bx + c = 0 is:

$x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

So, the first thing we're going to do is divide the whole function by 6:

6x^2 + 12x – 7 = 0 -> x^2 + 2x - 7/6

This step is optional, but it makes things quite easier.

Then we using the quadratic formula, where:

a=1, b= 2, c = -7/6.

Then:

$x=\frac{-2±\sqrt{2^{2}-4(1)(-7/6)}}{2}$

$x=\frac{-2±\sqrt{4 +14/3}}{2}$

$x=\frac{-2±\sqrt{26/3}}{2}$

So the zeros are:

$x1=\frac{-2+\sqrt{26/3}}{2}$

$x2=\frac{-2-\sqrt{26/3}}{2}$

4 0

2 years ago

This table represents a quadratic function. x y 0 -3 1 -3.75 2 -4 3 -3.75 4 -3 5 -1.75 What is the value of a in the function’s

serg [7]

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