Answer:
Given the equation: ![3x^2+10x+c =0](https://tex.z-dn.net/?f=3x%5E2%2B10x%2Bc%20%3D0)
A quadratic equation is in the form:
where a, b ,c are the coefficient and a≠0 then the solution is given by :
......[1]
On comparing with given equation we get;
a =3 , b = 10
then, substitute these in equation [1] to solve for c;
![x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%20%3D%20%5Cfrac%7B-10%5Cpm%20%5Csqrt%7B10%5E2-4%5Ccdot%203%20%5Ccdot%20c%7D%7D%7B2%20%5Ccdot%203%7D)
Simplify:
![x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%20%3D%20%5Cfrac%7B-10%5Cpm%20%5Csqrt%7B100-%2012c%7D%7D%7B6%7D)
Also, it is given that the difference of two roots of the given equation is
i.e,
![x_1 -x_2 = \frac{14}{3}](https://tex.z-dn.net/?f=x_1%20-x_2%20%3D%20%5Cfrac%7B14%7D%7B3%7D)
Here,
, ......[2]
.....[3]
then;
![\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B-10%20%2B%20%5Csqrt%7B100-%2012c%7D%7D%7B6%7D%20-%20%28%5Cfrac%7B-10%20%2B%20%5Csqrt%7B100-%2012c%7D%7D%7B6%7D%29%20%3D%20%5Cfrac%7B14%7D%7B3%7D)
simplify:
![\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%20%5Csqrt%7B100-%2012c%7D%20%7D%7B6%7D%20%3D%20%5Cfrac%7B14%7D%7B3%7D)
or
![\sqrt{100- 12c} = 14](https://tex.z-dn.net/?f=%5Csqrt%7B100-%2012c%7D%20%3D%2014)
Squaring both sides we get;
![100-12c = 196](https://tex.z-dn.net/?f=100-12c%20%3D%20196)
Subtract 100 from both sides, we get
![100-12c -100= 196-100](https://tex.z-dn.net/?f=100-12c%20-100%3D%20196-100)
Simplify:
-12c = -96
Divide both sides by -12 we get;
c = 8
Substitute the value of c in equation [2] and [3]; to solve ![x_1 , x_2](https://tex.z-dn.net/?f=x_1%20%2C%20x_2)
![x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}](https://tex.z-dn.net/?f=x_1%20%3D%20%5Cfrac%7B-10%20%2B%20%5Csqrt%7B100-%2012%5Ccdot%208%7D%7D%7B6%7D)
or
or
![x_1 = \frac{-10 + \sqrt{4}}{6}](https://tex.z-dn.net/?f=x_1%20%3D%20%5Cfrac%7B-10%20%2B%20%5Csqrt%7B4%7D%7D%7B6%7D)
Simplify:
![x_1 = \frac{-4}{3}](https://tex.z-dn.net/?f=x_1%20%3D%20%5Cfrac%7B-4%7D%7B3%7D)
Now, to solve for
;
![x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cfrac%7B-10%20-%20%5Csqrt%7B100-%2012%5Ccdot%208%7D%7D%7B6%7D)
or
or
![x_2 = \frac{-10 - \sqrt{4}}{6}](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cfrac%7B-10%20-%20%5Csqrt%7B4%7D%7D%7B6%7D)
Simplify:
![x_2 = -2](https://tex.z-dn.net/?f=x_2%20%3D%20-2)
therefore, the solution for the given equation is:
and -2.
C. 42.39 times 0.1 is 4.239. 42.39-4.239= 38.151 which rounds to 38.15
Answer:
1 to 25
Step-by-step explanation:
the actual person is 25 times the drawing
if Camille spent 19.40$ LESS than DOUBLE what Olivia spent then a rough estimate already can be Olivia spent 100 and Camille spent 172 but to find this out we have to add 19.40 onto 272.35 = 291.75 then we now that 291.75 is thrice what Olivia spent so if we third it we know what Olivia spent so 291.75/3 = 97.25 leaving 194.50 then if we minus 19.40 off 194.50 we get 175.10 so:
Camille spent: 175.10 and
Olivia Spent: 97.25
Answer:
(-1,1)
Step-by-step explanation:
All you have to do is plug it in the equation:
(x,y)
For (-1,1)
1=2(-1)+1
Simplify:
1=-1 but you know that 1≠-1 so (-1,1) is not it
For (3,7)
7=2(3)+1
Simplify:
7=7
For (-3,-5)
-5=2(-3)+1
-5=-5
Simplify:
For (0,1)
1=2(0)+1
1=1