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3 years ago

Bakit kinailangang umalis ng binatang Brahman sa kanilang bayan? Makatwiran ba ang

Mathematics

1 answer:

o-na [289]3 years ago

4 0

Answer:

Bakit kinailangang umalis ng binatang Brahman sa kanilang bayan? Makatwiran ba ang

kanyang paglisan? IpaliwanagBakit kinailangang umalis ng binatang Brahman sa kanilang bayan? Makatwiran ba ang

kanyang paglisan? Ipaliwanag

Step-by-step explanation:

Bakit kinailangang umalis ng binatang Brahman sa kanilang bayan? Makatwiran ba ang

kanyang paglisan? IpaliwanagBakit kinailangang umalis ng binatang Brahman sa kanilang bayan? Makatwiran ba ang

kanyang paglisan? Ipaliwanag

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Triangle XYZ is translated by the rule (x + 1, y − 1) and then dilated by a scale factor of 4 centered at the origin. Which stat

Doss [256]

Answer:

D) Segment YZ and segment Y"Z" are proportional after the dilation and congruent after the translation.

Step-by-step explanation:

After a dilation, a figure remains proportional because its angles and betweenness of points remain the same. After a translation, the image stays congruent to the pre-image.

3 0

3 years ago

Read 2 more answers

Let T: set of real numbers R Superscript nℝnright arrow→set of real numbers R Superscript mℝm be a linear transformation, and

Klio2033 [76]

Answer:

$\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent set.

Step-by-step explanation:

Given: $\{v_1,v_2,v_3\}$ is a linearly dependent set in set of real numbers R

To show: the set $\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent.

Solution:

If $\{v_1,v_2,v_3,...,v_n\}$ is a set of linearly dependent vectors then there exists atleast one $k_i:i=1,2,3,...,n$ such that $k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0$

Consider $k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0$

A linear transformation T: U→V satisfies the following properties:

1. $T(u_1+u_2)=T(u_1)+T(u_2)$

2. $T(au)=aT(u)$

Here, $u,u_1,u_2$ ∈ U

As T is a linear transformation,

$k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\$

As $\{v_1,v_2,v_3\}$ is a linearly dependent set,

$k_1v_1+k_2v_2+k_3v_3=0$ for some $k_i\neq 0:i=1,2,3$

So, for some $k_i\neq 0:i=1,2,3$

$k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0$

Therefore, set $\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent.

6 0

4 years ago

Oh mah lord please please help me

Alenkinab [10]

Option 2: $4^{\frac{1}{6}}$ is the correct answer.

Step-by-step explanation:

The radical expressions like these are simplified by using fractional exponents

given

$\frac{\sqrt{4}}{\sqrt[3]{4} }$

Converting radicals into exponents

When there is no base the exponent is 1/2 and as the base is 3, the exponent will be 1/3

$=\frac{4^{\frac{1}{2}}}{4^{\frac{1}{3}}}$

As the bases of numerator and denominator is same, the exponents can be subtracted

$=4^{\frac{1}{2}-\frac{1}{3}}\\=4^{\frac{3-2}{6}}\\=4^{\frac{1}{6}}$

Hence,

Option 2: $4^{\frac{1}{6}}$ is the correct answer.

Keywords: Exponents, radicals

Learn more about radicals at:

#LearnwithBrainly

7 0

3 years ago

At a factory, Taj produces 23 toys per hour and Rosa produces 47 toys per hour. How many hours will it take for

Kruka [31]

Answer:

7 hours

Step-by-step explanation:

47+23=70

1hour = 70 toys

? = 490 toys

Then you cross multiply:

1hour*490 toys= 490/70 = 7 hours

____________

70 toys

8 0

3 years ago

There is another way to rank the difficulty of test questions using Networks: Each student gives scores for each problem, then w

Vlad [161]

Answer:

Step-by-step explanation:

Truly, on a single student level, the Massey method can be applied i.e. problems are ranked according to their level of difficulty for a single student.

Let us say there are 4 problems P1, P2, P3, P4 set in an exam where a student has to solve any two.

Assuming student 1 solves and scores more in P1 than in P2, r1>r2

student 2 more in P1 than P4 r1>r4

student 3 more in P2 than P4 r2>r4

student 4 more in P4 than P3 r4>r3

student 5 more in P2 than P3 r2>r3

and such many more cases.

then the Massey's model can be written as

P1 P2 P3 P4

\begin{bmatrix} 1 &-1 &0 &0\\ 1 &-1 &0 &0 \\ 0 &1 & 0 &-1 \\ 0 & 0 &-1 & 1 \\ 0 & 1 & -1 & 0 \end{bmatrix}.

\times \begin{bmatrix}r1 \\r2 \\r3 \\r4 \end{bmatrix} =

\begin{bmatrix}y_{1} \\ y_{2} \\ y_{3} \\ y_{4}\\ y_{5}\end{bmatrix}.

If Pi gives more score than Pj, the entry below Pi shall be 1 and below Pj shall be -1.

The rest of the entries shall be 0.

r1,r2,r3,r4,r5 are ranks .

There may be more rows depending upon more combination of 2 problems.

Should in case more than one student solve the same combination and get scores in the same order, for instance, 12 students solve P1 and P2, for 7 of them r1>r2, then their scores can be averaged.

Normally, such systems are overdetermined, i.e., more rows than the number of unknowns, then it is solved by the Least square method ., as there shall be no a solution with least error is found out.

The overall ranking, in reverse order, shall give the ranking according to increasing difficulty.

7 0

4 years ago