Answer:
See picture and explanation below.
Step-by-step explanation:
With this information, the matrix A that you can find is the transformation matrix of T. The matrix A is useful because T(x)=Av for all v in the domain of T.
A is defined as ![T=([T(v_1)] [T(v_2] [T(v_3)])\text{ where }[T(v_i)]](https://tex.z-dn.net/?f=T%3D%28%5BT%28v_1%29%5D%20%5BT%28v_2%5D%20%5BT%28v_3%29%5D%29%5Ctext%7B%20where%20%7D%5BT%28v_i%29%5D)
denotes the vector of coordinates of 
respect to the basis 
(we can apply this definition because 
forms a basis for the domain of T).
The vector of coordinates can be computed in the following way: if 
then ![[T(v_i)]=(a_1,a_2,a_3)^t](https://tex.z-dn.net/?f=%5BT%28v_i%29%5D%3D%28a_1%2Ca_2%2Ca_3%29%5Et)
.
Note that we have all the required information: 
then ![[T(v_1)]=(0,1,0)^t](https://tex.z-dn.net/?f=%5BT%28v_1%29%5D%3D%280%2C1%2C0%29%5Et)
hence ![[T(v_2)]=[T(v_3)]=(1,0,0)^t](https://tex.z-dn.net/?f=%5BT%28v_2%29%5D%3D%5BT%28v_3%29%5D%3D%281%2C0%2C0%29%5Et)
The matrix A is on the picture attached, with the multiplication A(1,1,1).
Finally, to obtain the output required at the end, use the properties of a linear transformation and the outputs given:
In this last case, we can either use the linearity of T or multiply by A.
At x=0 y=6 the y intercept. The rate is -0.25 so the equation is y=6-0.25x. The ordered pair is (x,6-0.25x).
Answer:
d=4.5t
Step-by-step explanation:
in two hours, Francine went 9 miles
9/2=4.5
the distance equals 4.5 times each cycling time.
d=4.5t
Answer: Iron
Explanation:
The food contain high iron food as the daily value (DV) of the iron is 45% and most of the nutrient claimed that food contain low sodium but contain good source of the fiber. The food is high in the amount of the nutrient with 20% daily value.
The daily value actually recommend the minimum amount of the particular nutrient which everyone should consume in daily life.
11. You’ve done it correctly
12. Let x^2=y
y^2+13y+40=0
(y+8)(y+5)=0
y=8, 5
Since y=x^2
x^2=8 x^2=5
x=+/-√5 x= +/-2√2
13. x^4-x^2-x^2-8=0
x^4-2x^2-8=0
let x^2=y
Y^2-2y-8=0
(y-4)(y+2)=0
y=4, -2
Since y=x^2
X^2=4 X^2=-2
X= +/- 2 This wouldn’t be a real solution
14. It’s pretty much the same process, just substitute y in for x^2. If you’re confused feel free to ask and I can do it, or you can put it through Photomath
15. You’re on the right track so I’m just going to continue from where you left off
x^2(4x+5)-4(x+5)=0
(x^2-4)(4x+5)=0
x= +/- 2 4x=5
x=5/4 or 1 1/4
Hope this helped :)
