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RideAnS [48]
2 years ago
6

Helpppppp meeee please and thank you

Mathematics
2 answers:
shusha [124]2 years ago
8 0

Answer:

c

Step-by-step explanation:

labwork [276]2 years ago
3 0

Answer It would be D

Step-by-step explanation:

Because it is isosolies M is mediocer so it is option D

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Find the domain and range pls!
skad [1K]
Domain should be, (-♾️,♾️)
And range should be, (-♾️, 2)

If you ment something other then this please message me so I can help!
4 0
2 years ago
Which of the following subsets of ℝ3×3 are subspaces of ℝ3×3? A. The 3×3 matrices whose entries are all greater than or equal to
Debora [2.8K]

Answer:

A. It is NOT a subspace of R^3x3

B. It IS a subspace of R^3x3

C. It IS a subspace of R^3x3

D. It is NOT a subspace of R^3x3

E  It is NOT a subspace of R^3x3

F.  It IS a subspace of R^3x3

Step-by-step explanation:

A way to show that a set is not a subspace, it´s enough to show that some properties of the definition of a vector spaces does not apply in that set or that operations under that set are not closed (we can get out of the set with linear combinations of elements in the set).

A. For definition of subspace, we know that every element has to have an additive inverse, but in set "A" (The 3×3 matrices whose entries are all greater than or equal to 0 ) every entry is greater than or equal to zero. In this set, there´s no additive inverse with the usual sum in R^3x3.

If sufficient to prove a set is a subspace showing that zero is in the set, there are additive inverses and that operations (sum and scalar multiplication) are closed in that set.

B.  Notice that the matrix 0 is in "B" (The 3×3 matrices A such that the vector (276) is in the kernel of A), also notice if A(276)=0 then -A(276)=0 so every additive inverse (of an element in "B") belongs to "B".

Now we just have to prove that operations are closed in "B". Let X,Y matrices in set "B" and let z a scalar from the field. We are going to show that:

zX+Y ∈ B

For definition of set B:

X(276)=0 and Y(276)=0

So for zX+Y:

(zX+Y)(276)=zX(276)+Y(276)=z(0)+(0)

(zX+Y)(276)=0

So (276) is in the kernel of zX+Y, i.e (zX+Y) ∈ B.

We conclude "B" (with usual sum and scalar product of matrices) is a subspace of R^3x3

C. Notice the matrix 0 ∈ "C" (The diagonal 3×3 matrices) and there are all the additive inverse of the elements in "C". With the usual sum and scalar product, if the only zero entries are above and under the diagonal, it´ll stay like that no matter what linear combination we do because sum of matrices is entry by entry, and for every entry above or under the diagonal the sum and scalar product of two elements is going to be 0 in the same entries under and above the diagonal. "C" is a subspace

D.  In set "D" (The non-invertible 3×3 matrices) it´s necessary to show that the sum is not closed:

Consider the following matrices and their sum:

X=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&0\end{array}\right]\\ Y=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&1\end{array}\right]

X+Y=I

We showed that sum is not closed in "C", so "C" is not a subspace of R^3x3

E. The definition of a reduced row-echelon matrix requires that the first element of a row must be 1, but with sum and scalar multiplication is easy to show that these pivot could easily change its value. So the set "E" is not closed under the usual operations under R^3x3.

F. The argument is similar to part C. No matter what linear combination we do, the last row is always going to be zero (with the usual operations in R^3x3). 0 ∈ "F" (The 3×3 matrices with all zeros in the third row) and all additive inverses (for an element in "F") is in "F", we affirm that "F" is a subspace of R^3x3

5 0
2 years ago
HELP ASAP!!!!!<br><br> 2x 2y=5 <br> y=1/4x
andrey2020 [161]

Answer:

X=2

Y=1/4*2

Y=1/2

Y=.5

Step-by-step explanation:

PROOF

2*2+2*.5=5

4+1=5

5=5

3 0
3 years ago
The time it takes to travel a fixed distance varies inversely with the speed traveled. It takes
uysha [10]

first off, let's notice that Purple's time is in minutes, whilst the rate is in miles per hour, the units of both must correspond, so, we can either change the time from minutes to hours or the rate from hours to minutes, hmmm let's change the time to hours.

so 40 minutes, we know there are 60 minutes in 1 hour, so 40 minutes will be 40/60 of an hr, or namely 2/3.

\qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}} ~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\cfrac{\stackrel{\downarrow }{k}}{x}~\hfill } \\\\ \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] ~\dotfill

\stackrel{\begin{array}{llll} \textit{\tiny "t"ime varies}\\ \textit{\tiny inversely with "s"peed} \end{array}}{t = \cfrac{k}{s}}\qquad \textit{we know that} \begin{cases} t=\stackrel{minutes}{40}\to \stackrel{hrs}{\frac{2}{3}}\\ s=\stackrel{m/h}{9} \end{cases} \implies \cfrac{2}{3}~~ = ~~\cfrac{k}{9} \\\\\\ 18=3k\implies \cfrac{18}{3}=k\implies 6=k~\hfill \boxed{t=\cfrac{6}{s}} \\\\\\ \textit{when s = }\stackrel{m/h}{12}\textit{ what is "t"?}\qquad t=\cfrac{6}{12}\implies t=\cfrac{1}{2}

6 0
2 years ago
You have $138 in a savings account and save $12 per week. Your friend has $98 in a savings account and saves $17 per week. How m
r-ruslan [8.4K]

Solution

Provided :-

• I have $138

• I save $12 per week

• Friend have $ 98

• He saves $17 per week

Let the number of weeks required to have same balance be x

→ 138 + 12x = 98 + 17x

→ 138 - 98 = 17x - 12x

→ 40 = 5x

→ x = 8

So we need 8 weeks

4 0
2 years ago
Read 2 more answers
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