Funtion ! in vertex form is given by
<span>f(x) = 4x^2 + 8x + 1</span> = 4(x^2 + 2x + 1/4) = 4(x^2 + 2x + 1 + 1/4 - 1) = 4(x + 1)^2 + 4(-3/4) = 4(x + 1)^2 - 3
Thus, the least minimun value is (-1, -3)
Also, the least minimum value of function 2 is (-1, 0)
Therefore, function 1 has the least minimum value at (-1, -3)
Answer:
1)a
2)d
3)b
4)c
Step-by-step explanation:
We have to find the mass of the gold bar.
We have gold bar in the shape of a rectangular prism.
The length, width, and the height of the gold bar is 18.00 centimeters, 9.21 centimeters, and 4.45 centimeters respectively.
First of all we will find the volume of the gold bar which is given by the volume of rectangular prism:
Volume of the gold bar 
Plugging the values in the equation we get,
Volume of the gold bar 
Now that we have the volume we can find the mass by using the formula,

The density of the gold is 19.32 grams per cubic centimeter. Plugging in the values of density and volume we get:
grams
So, the mass of the gold bar is 14252.769 grams
Answer:
1) False
2) False
3) True
4) False
Step-by-step explanation:
1) Flase, {v1,v2,v3, ..., vp} is a base for H when they span H and also they are linearly independent.
2) False. A single nonzero vector is linearly independent , not dependent. There is not null linear combination that gives 0 as a result involving that vector.
3) True, if the columns werent linearly independent, we could triangulate the matrix and obtain 0, so the matrix wouldnt be invertible. This means that the columns should be linearly independent for the matrix to be invertible and as a consecuence, they will spam a subspace of R^n of dimension n, which means that they will spam all R^n and therefore, they form a basis of R^n.
4) False. A basis is a spanning set that is as small as possible. Larger spanning sets will have extra elements apart from those who can form a base toguether. Those elements will make the set linearly dependent.
Answer:
It took 1.2 hours to get to the store
Step-by-step explanation:
Let the time taken to reach the store be t₁
Let the time taken to come back be t₂
Let the speed to and from store = s₁ and s₂ respectively
let the distance to the store = d
To the store:

Back from the store:

We are told that total time (t₁ + t₂) = 2 hours
t₁ + t₂ = eqn (1) + eqn (2)

∴ length of trip to the store = t₁
from eqn (1)
