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2 years ago

Let:T : ℝ3→ℝ3 be the transformation that projects each vector x=(x1,x2,x3)onto the plane

Mathematics

1 answer:

dangina [55]2 years ago

5 0

Answer:

Step-by-step explanation:

A linear transformation must satisfy the following properties.

- T(0) = 0.

- For vector a,b then T(a+b) = T(a) + T(b).

- For a vector a and a scalar r, it must happen that T(ra) = rT(a)

In this case we have that T(a,b,c) = (a,0,c).

Note that T(0) = T(0,0,0) = (0,0,0) = 0. So, the first property holds.

Let $a=(a_1,a_2,a_3), b=(b_1,b_2,b_3)$ . Then

$T(a+b) = T((a_1+b_1,a_2+b_2,a_3+b_3)) = (a_1+b_1,0,a_3+b_3) = (a_1,0,a_3)+(b_1,0,b_3) = T(a) + T(b)$

So the second property holds.

Finally, let r be a scalar and let $a=(a_1,a_2,a_3)$ . Then

$T(ra) = T((ra_1,ra_2,ra_3)) = (ra_1,0,ra_3) = r(a_1,0,a_3)= rT(a)$

So, the three properties hold, and therefore, T is a linear transformation.

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ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl

melamori03 [73]

Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).

Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

$\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x} \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm$

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

$\frac{AQ}{QP}=\frac{AB}{CP} \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19$

$\frac{BQ}{CQ}=\frac{AB}{CP} \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8$

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

7 0

3 years ago

Suppose the speeds of cars along a stretch of I-40 is normally distributed with a mean of 70 mph and standard deviation of 5 mph

neonofarm [45]

Answer:

a) 68%.

b) 84%.

c) Approximately 2.5% of cars are traveling at a speed greater than or equal to 80 mph.

d) Approximately 2.5% of cars are traveling at a speed greater than or equal to 80 mph.

e) Between 60 mph and 80 mph.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean of 70 mph, standard deviation of 5 mph.

(a) Approximately what percent of cars are travelling between 65 and 75 mph?

70 - 5 = 65

70 + 5 = 75

Within 1 standard deviation, so approximately 68%.

(b) If the speed limit on this stretch of highway is 65 mph, approximately what percent of cars are traveling faster than the speed limit?

The normal distribution is symmetric, which means that 50% of the measures are below the mean, and 50% are above.

65 is one standard deviation below the mean, so of the cars below the mean, 68% are above 65 mph.

0.68*50% + 50% = 34% + 50% = 84%.

84%.

(c) What percent of cars are traveling at a speed greater than or equal to 80 mph?

80 = 70 + 2*10

2 standard deviations above the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean. Due to the symmetry of the normal distribution, of the other 5%, 2.5% is at least 2 standard deviations below the mean and 2.5% is at least 2 standard deviations above the mean. Then:

Approximately 2.5% of cars are traveling at a speed greater than or equal to 80 mph.

(d) What percent of cars are traveling at a speed greater than 80 mph?

Same as item c, as in the normal distribution, the probability of an exact value is considered to be 0.

(e) 95% of cars are traveling between what two speeds?

Within two standard deviations of the mean.

70 - 2*5 = 60 mph

70 + 2*5 = 80 mph.

Between 60 mph and 80 mph.

6 0

2 years ago

Find the quotient: -10/19 divided by (-5/7) ?

zhannawk [14.2K]