All categories

3 years ago

HELP PLEASE! 50 POINTS AND WILL MARK BRAINLIEST.

Mathematics

1 answer:

Gelneren [198K]3 years ago

3 0

Answer:

I know looks confusing

Step-by-step explanation:

f(1)=35f(n)=f(n−1)−2f(2)=f(1)−2=35−2=33f(3)=f(2)−2=33−2=31f(4)=f(3)−2=31−2=29f(5)=f(4)−2=29−2=27

Answer: 35, 33, 31, 29, 27.

You might be interested in

Assume that T is a linear transformation. Find the standard matrix of T. T: R^3 right arrow R^2 , T(e 1) =(1,2), and T(e2 ) =( -

irina [24]

Answer:

$A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]$

Step-by-step explanation:

Given

$T:R^3->R^2$

$T(e_1) = (1,2)$

$T(e_2) = (-4,6)$

$T(e_3) = (2,-6)$

Required

Find the standard matrix

The standard matrix (A) is given by

$Ax = T(x)$

Where

$T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]$

$Ax = T(x)$ becomes

$Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]$

The x on both sides cancel out; and, we're left with:

$A = [T(e_1)\ T(e_2)\ T(e_3)]$

Recall that:

$T(e_1) = (1,2)$

$T(e_2) = (-4,6)$

$T(e_3) = (2,-6)$

In matrix:

$(a,b)$ is represented as: $\left[\begin{array}{c}a\\b\end{array}\right]$

So:

$T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]$

$T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]$

$T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]$

Substitute the above expressions in $A = [T(e_1)\ T(e_2)\ T(e_3)]$

$A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]$

Hence, the standard of the matrix A is:

$A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]$

6 0

3 years ago

The owner of a small restaurant bought 75 kilograms of rice. Each week, the restaurant uses 4.5 kilograms of rice.

leva [86]

Answer:

6 weeks 27kilo

12 weeks 54 kilo

Step-by-step explanation:

4.5*12=54

4.5*6=27

3 0

3 years ago

HEEELLLPPP!!!!!!!!! PLEEAASEEE!!!!!!!!

Ksivusya [100]

Answer:

Step-by-step explanation:

$17. (a^{m})^{n}=a^{m*n}\\\\a^{m}*a^{n}=a^{m+n}\\\\-3k^{2}*(4k^{5})^{3}=-3k^{2}*4^{3}*k^{5*3}\\\\ = -3k^{2}*64*k^{15}\\\\= (-3)*64*k^{2+15}\\\\= - 192k^{17}$

$19. \dfrac{a^{m}}{a^{n}}=a^{m-n}\\\\\\\dfrac{(-7p^{4})*(-8p^{5})}{2p^{3}}= \dfrac{(-7)*(-8)*p^{4}*p^{5}}{2p^{3}}\\\\\\=(-7)*(-4)*p^{4+5-3} \\\\= 28p^{6}\\$

$20)\dfrac{15a^{16}b^{11}}{(3a^{4}b^{2})^{3}}=\dfrac{15a^{16}b^{11}}{3^{3}*a^{4*3}*b^{2*3}}\\\\\\=\dfrac{15a^{16}b^{11}}{27*a^{12}*b^{6}}\\\\=\dfrac{5*a^{16-12}*b^{11-6}}{9}\\\\\\=\dfrac{5a^{4}b^{5}}{9}$

4 0

2 years ago

Use the point-slope form linear equation given.

Agata [3.3K]

y-3=3(x+1)

opening the bracket

y-3=3x+3

y=3x+3+3

equation of the line in the form y=mx+c;

y=3x+6

therefore gradient=3

parallel lines have same gradient therefore gradient of the other line is 3

y--3/x-0=3

y+3=3(x-0)

y+3=3x-0

y=3x-3.

8 0

3 years ago

I NEED HELP PEOPLE!!!! WILL GIVE BRAINLIEST!!!!

Sauron [17]

Answer:

L(2,4) M(3,1) N(0,4)

Step-by-step explanation:

6 0

3 years ago