For the 2nd part of 2. just plug in what you have for G in your previous graph into the equation. This will give you H for all 5 columns . Like 3×8(-1+5)= h = 3× 32= 96 so H should equal 96 and so on as far as this function.
For number 3. The equation is given so just plug in your T for time which is 3 seconds, so...-16(3)^2+90(3) = H the height at 3 seconds. I'm doing it in my head but should be the height is 414. You should also say whether it's ft or inches etc because the teacher or yourself left that out of the equation which is also vital lol.
Answer:
is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.
Step-by-step explanation:
Given the function
![f\left(x\right)=x^3-6x^2+3x+10](https://tex.z-dn.net/?f=f%5Cleft%28x%5Cright%29%3Dx%5E3-6x%5E2%2B3x%2B10)
As the highest power of the x-variable is 3 with the leading coefficients of 1.
- So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.
solving to get the zeros
![f\left(x\right)=x^3-6x^2+3x+10](https://tex.z-dn.net/?f=f%5Cleft%28x%5Cright%29%3Dx%5E3-6x%5E2%2B3x%2B10)
∵ ![f(x)=0](https://tex.z-dn.net/?f=f%28x%29%3D0)
as
![Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0](https://tex.z-dn.net/?f=Factor%5C%3Ax%5E3-6x%5E2%2B3x%2B10%5C%3A%3A%5C%3A%5Cleft%28x%2B1%5Cright%29%5Cleft%28x-2%5Cright%29%5Cleft%28x-5%5Cright%29%3D0)
so
Using the zero factor principle
if ![ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)](https://tex.z-dn.net/?f=ab%3D0%5C%3A%5Cmathrm%7Bthen%7D%5C%3Aa%3D0%5C%3A%5Cmathrm%7Bor%7D%5C%3Ab%3D0%5C%3A%5Cleft%28%5Cmathrm%7Bor%5C%3Aboth%7D%5C%3Aa%3D0%5C%3A%5Cmathrm%7Band%7D%5C%3Ab%3D0%5Cright%29)
![x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0](https://tex.z-dn.net/?f=x%2B1%3D0%5Cquad%20%5Cmathrm%7Bor%7D%5Cquad%20%5C%3Ax-2%3D0%5Cquad%20%5Cmathrm%7Bor%7D%5Cquad%20%5C%3Ax-5%3D0)
![x=-1,\:x=2,\:x=5](https://tex.z-dn.net/?f=x%3D-1%2C%5C%3Ax%3D2%2C%5C%3Ax%3D5)
Therefore, the zeros of the function are:
![x=-1,\:x=2,\:x=5](https://tex.z-dn.net/?f=x%3D-1%2C%5C%3Ax%3D2%2C%5C%3Ax%3D5)
is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.
Therefore, the last option is true.
X=7
100-30=70
70/10
x=7
hope this helps
Answer:
(-3,5) Quad 2
(4,-1) Quad 4
(2,0) Positive X axis
(2,2) Quad 1
(-3,-6) Quad 3
Step-by-step explanation:
Answer:
Option B.
Step-by-step explanation:
Consider the below figure attached with this question.
It is given that triangle ABC is an isosceles triangle.
From the below figure it is clear that the vertices of triangle are A(-2,-4), B(2,-1) and C(3,-4).
Distance formula:
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
Using this formula, we get
![AB=\sqrt{(2-(-2))^2+(-1-(-4))^2}=\sqrt{16+9}=5](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%282-%28-2%29%29%5E2%2B%28-1-%28-4%29%29%5E2%7D%3D%5Csqrt%7B16%2B9%7D%3D5)
![BC=\sqrt{(3-2)^2+(-4-(-1))^2}=\sqrt{1+9}=\sqrt{10}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B%283-2%29%5E2%2B%28-4-%28-1%29%29%5E2%7D%3D%5Csqrt%7B1%2B9%7D%3D%5Csqrt%7B10%7D)
![AC=\sqrt{(3-(-2))^2+(-4-(-4))^2}=\sqrt{25}=5](https://tex.z-dn.net/?f=AC%3D%5Csqrt%7B%283-%28-2%29%29%5E2%2B%28-4-%28-4%29%29%5E2%7D%3D%5Csqrt%7B25%7D%3D5)
Now,
Perimeter of triangle ABC = AB + BC + AC
![=5+\sqrt{10}+5](https://tex.z-dn.net/?f=%3D5%2B%5Csqrt%7B10%7D%2B5)
![=10+\sqrt{10}](https://tex.z-dn.net/?f=%3D10%2B%5Csqrt%7B10%7D)
So, perimeter of triangle ABC is
units.
Therefore, the correct option is B.