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3 years ago

If a-2= (2^2/3+2^1/3) find a^3-6a^2+12a-14

Mathematics

2 answers:

raketka [301]3 years ago

7 0

Answer:

Step-by-step explanation:

7. 1, for r = 0 - 1, for r = 1 Hence, determine alo. Using characteristic root ... find the solution of the recurrence relation y, + 9 y, 2 = 6y, 1, subjected to the ... Solve a, -5a, 1 + 6a, 2 = 0 , given initial conditions ao = 2 and a1 = 5. ... Solve the recurrence relation a, – 7a, 1 + 16a, 2 – 12a, 3 = 0 for n > 3 with ... 2"; 3. a = (2)” – n.

Talja [164]3 years ago

4 0

Answer:

Step-by-step explanation:

I solved in the picture

Hope this helps ^-^

You might be interested in

The average human gestation period is 270 days with a standard deviation of 9 days. The period is normally distributed. What is

Anna35 [415]

Answer:

Probability that a randomly selected woman's gestation period will be between 261 and 279 days is 0.68.

Step-by-step explanation:

We are given that the average human gestation period is 270 days with a standard deviation of 9 days. The period is normally distributed.

Firstly, Let X = women's gestation period

The z score probability distribution for is given by;

Z = $\frac{ X - \mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = average gestation period = 270 days

$\sigma$ = standard deviation = 9 days

Probability that a randomly selected woman's gestation period will be between 261 and 279 days is given by = P(261 < X < 279) = P(X < 279) - P(X $\leq$ 261)

P(X < 279) = P( $\frac{ X - \mu}{\sigma}$ < $\frac{279-270}{9}$ ) = P(Z < 1) = 0.84134

P(X $\leq$ 261) = P( $\frac{ X - \mu}{\sigma}$ $\leq$ $\frac{261-270}{9}$ ) = P(Z $\leq$ -1) = 1 - P(Z < 1)

= 1 - 0.84134 = 0.15866

Therefore, P(261 < X < 279) = 0.84134 - 0.15866 = 0.68

Hence, probability that a randomly selected woman's gestation period will be between 261 and 279 days is 0.68.

3 0

3 years ago

Instructions:Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17). ∆XYZ has

Akimi4 [234]

to compare the triangles, first we will determine the distances of each side

Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
Solving

∆ABC A(11, 6), B(5, 6), and C(5, 17)

AB = 6 units BC = 11 units AC = 12.53 units
∆XYZ X(-10, 5), Y(-12, -2), and Z(-4, 15)
XY = 7.14 units YZ = 18.79 units XZ = 11.66 units

∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).

MN = 6 units NO = 11 units MO = 12.53 units
∆JKL J(17, -2), K(12, -2), and L(12, 7).
JK = 5 units KL = 9 units JL = 10.30 units
∆PQR P(12, 3), Q(12, -2), and R(3, -2)
PQ = 5 units QR = 9 units PR = 10.30 units
Therefore
we have the ∆ABC and the ∆MNO 
with all three sides equal ---------> are congruent
we have the ∆JKL and the ∆PQR
with all three sides equal ---------> are congruent

let's check

Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

1) If ∆MNO ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC

then ∆MNO ≅ ∆ABC

a) Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).

M(-9, -4)-----------------> M1(-9,4)

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

b) Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

∆M1N1O1 M1(-9, 4), N1(-3, 4), and O1(-3, 15).

M1(-9, -4)-----------------> M2(9,4)

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

c) Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

∆M2N2O2 M2(9,4), N2(3,4), and O2(3, 15).

M2(9, 4)-----------------> M3(11,6)=A

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

∆ABC A(11, 6), B(5, 6), and C(5, 17)

∆MNO reflection------- > ∆M1N1O1 reflection---- > ∆M2N2O2 translation -- --> ∆M3N3O3

The ∆M3N3O3=∆ABC

Therefore ∆MNO ≅ ∆ABC - > check list

2) If ∆JKL -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR

then ∆JKL ≅ ∆PQR

d) Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

∆JKL J(17, -2), K(12, -2), and L(12, 7).

J(17, -2)-----------------> J1(2,17)

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

e) translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

∆J1K1L1 J1(2, 17), K1(2, 12), and L1(-7, 12).

J1(2, 17)-----------------> J2(12,3)=P

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

∆PQR P(12, 3), Q(12, -2), and R(3, -2)

∆JKL rotation------- > ∆J1K1L1 translation -- --> ∆J2K2L2=∆PQR

Therefore ∆JKL ≅ ∆PQR - > check list

6 0

3 years ago

The given triangles are similar. Solve for x.

Maru [420]

The first one is 21.93 rounded up to 22.

8 0

3 years ago

Please Evaluate 27 times ( 1/3) to the 3 power. A). 1 B). 3 C). 9 D). 27

topjm [15]

Answer:

you want to follow PEMDAS so you would multiply 27 by 1/3 to get 81.003, which you would round to 81, then you would multiply 8 to the third power and you would get 512.

Step-by-step explanation:

27(1/3)^3

81^3

512

7 0

3 years ago

Ahsan

NikAS [45]

It is number 1 and number 2 because the equation is 242× 62 and Number 1 and 2 are just breaking it down

7 0

3 years ago

If a-2= (2^2/3+2^1/3) find a^3-6a^2+12a-14​

If a-2= (2^2/3+2^1/3) find a^3-6a^2+12a-14