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4 years ago

7

Oblicz granicę ciągu (pierwiastek z (n^2+6)/2n+2).

1 answer:

kap26 [50]4 years ago

6 0

Answer:

umm what

Step-by-step explanation:

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If the length of Mike’s rectangular poster is 2 times its width. If the perimeter is 12 inches, what is the area of the poster?

seraphim [82]

Answer:

Area of Poster = 23 sq inches

Step-by-step explanation:

Let say Width of Rectangular Poster = W inches

then Length of Rectangular Post = 2W inches

Perimeter = 2 ( W + 2W) = 6W inches

Perimeter given = 12 inches

Equating both

6W = 24

=> W = 4

Width of Poster = W = 4 Inches

Length of Poster = 2W = 2 * 4 = 8 inches

Area of Poster = 4 * 8 = 32 sq inches

7 0

3 years ago

In triangle ABC shown below, side AB is 8 and side AC is 6: Triangle ABC with segment joining point D on segment AB and point E

RSB [31]

For the segment DE to be the midsegment of the triangle ABC, the point D must cut AB in two equal halves, and point E must cut AC in two equal halves as well, that way, DE touches both ends right in the middle

and that can only occur if AD is 4 (half of 8) and AE is 3 ( half of 6)

7 0

3 years ago

Read 2 more answers

Enter a number.<br> The measure of the hypotenuse is __________

poizon [28]

Answer:

Step-by-step explanation:

c^2 = 24^2 + 7^2

c^2 = 576 + 49

c^2 = 625

c = √625

c = 25

4 0

4 years ago

There are 3 teachers to 125 pupils during the school program. How many teachers were

Kobotan [32]

Answer:

There were 60 teachers.

Step-by-step explanation:

☆We can use proportion to solve this!

$\frac{3 \: teachers}{125 \: pupils} = \frac{x}{2500 \: pupils} \\ \\ \frac{125x}{125} = \frac{7500}{125} \\ \\ x = 60$

5 0

3 years ago

Suppose that T : R3 → R2 is given by:

Ad libitum [116K]

Answer: The required answers are

(a) T is proved to be a linear transformation.

(b) The matrix A such that T(x) = Ax is $\begin{pmatrix}1 & 0 &0 \\ 0 & 1 &0 \end{pmatrix}$

Step-by-step explanation: We are given a linear transformation T : R³ → R² defined as follows :

$T(a,b,c)=(a,b).$

We are to

(a) prove that T is a linear transformation

and

(b) find a matrix A such that T(x) = Ax.

(a) Let s, t are any real numbers and (a, b, c), (a', b', c') ∈ R³.

Then, we have

$T(s(a,b,c)+t(a',b',c'))\\\\=T(sa+ta',sb+tb',sc+tc')\\\\=(sa+ta',sb+tb')\\\\=(sa,sb)+(ta'+tb')\\\\=s(a,b)+t(a',b')\\\\=sT(a,b,c)+tT(a',b',c').$

So, we get

$T(s(a,b,c)+t(a',b',c'))=sT(a,b,c)+tT(a',b',c').$

Therefore, T is a linear transformation.

(b) We know that B = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a standard basis for R³ and B' = {(1, 0), (0, 1)} is a standard basis for R².

So, we have

$T(1,0,0)=(1,0)=1(1,0)+0(0,1),\\\\T(0,1,0)=(0,1)=0(1,0)+1(0,1),\\\\T(0,0,1)=(0,0)=0(1,0)+0(0,1).$

So, the matrix A such that T(x) = Ax will be given by

$\begin{pmatrix}1 & 0 &0 \\ 0 & 1 &0 \end{pmatrix}$

Thus,

(a) T is proved to be a linear transformation.

(b) The matrix A such that T(x) = Ax is $\begin{pmatrix}1 & 0 &0 \\ 0 & 1 &0 \end{pmatrix}$

4 0

3 years ago