What is the prime factoriztion 153

1. The most massive planet in the solar system, Jupiter, has a mass of about 1,898,000,000,000,000,000,000,000,000 kg. Estimate

UNO [17]

Answer:

D. $2*10^{27} kg$

Step-by-step explanation:

Mass of Jupiter is given as 1,898,000,000,000,000,000,000,000,000 kg

Estimating this to the nearest whole number would be = 2,000,000,000,000,000,000,000,000,000 kg (rounding up the the "1,898")

To express this in scientific notation form, simply count how many zeros we have to the nearest whole number, 2.

We have 27 zeros there.

Therefore, we would multiply the whole number, 2, by 10 raise to the power of 27.

An estimate of the mass of Jupiter = $2*10^{27} kg$

6 0

3 years ago

Let v = (v1, v2) be a vector in R2. Show that (v2, −v1) is orthogonal to v, and use this fact to find two unit vectors orthogona

andrey2020 [161]

Answer:

a. v.v' = v₁v₂ - v₁v₂ = 0 b. (20, -21)/29 and (-20,21)/29

Step-by-step explanation:

a. For two vectors a, b to be orthogonal, their dot product is zero. That is a.b = 0.

Given v = (v₁, v₂) = v₁i + v₂j and v' = (v₂, -v₁) = v₂i - v₁j, we need to show that v.v' = 0

So, v.v' = (v₁i + v₂j).(v₂i - v₁j)

= v₁i.v₂i + v₁i.(- v₁j) + v₂j.v₂i + v₂j.(- v₁j)

= v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

i.i = 1, i.j = 0, j.i = 0 and j.j = 1

So, v.v' = v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

= v₁v₂ × 1 - v₁v₁ × 0 + v₂v₂ × 0 - v₂v₁ × 1

= v₁v₂ - v₂v₁

= v₁v₂ - v₁v₂ = 0

So, v.v' = 0

b. Now a vector orthogonal to the vector v = (21,20) is v' = (20,-21).

So the first unit vector is thus a = v'/║v'║ = (20, -21)/√[20² + (-21)²] = (20, -21)/√[400 + 441] = (20, -21)/√841 = (20, -21)/29.

A unit vector perpendicular to a and parallel to v is b = (-21, -20)/29. Another unit vector perpendicular to b, parallel to a and perpendicular to v is thus a' = (-20,-(-21))/29 = (-20,21)/29

8 0

3 years ago

If you roll two fair dice repeatedly, what is the probability that you will get a sum of 4 before you get a sum of 5 ? (a) (b) (

AlexFokin [52]

Answer:

The probability that you will get a sum of 4 before you get a sum of 5 is $\frac{1}{12}$

Step-by-step explanation:

Given : If you roll two fair dice repeatedly.

To find : What is the probability that you will get a sum of 4 before you get a sum of 5 ?

Solution :

When two dice are rolled the outcomes are

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Total number of outcomes = 36

Favorable outcome get a sum of 4 before you get a sum of 5 is (1,3) ,(2,2) and (3,1) = 3

The probability that you will get a sum of 4 before you get a sum of 5 is

$P=\frac{3}{36}$

$P=\frac{1}{12}$

Therefore, The probability that you will get a sum of 4 before you get a sum of 5 is $\frac{1}{12}$

5 0

3 years ago

David went to a fast-food shop for a meal in a late evening. Only steak, pork chop and grilled chicken were available for main c

Doss [256]

six

Step-by-step explanation:

the reason is this: let steak be (s), pork be (p), grilled chicken be (gc), tea be (t), and coffee be (c)

now, we want to find the total number of outcomes, and this cannot be determined by a simple roll of the dice. you have to count the sample space required for EVERY outcome, and it has to lie within the conditions. so, assuming David only takes 1 main course and 1 drink, he would be down to six option, which are :

1. s & t

2. s & c

3. p & t

4. p & c

5. gc & t

6. gc & c

5 0

3 years ago

Help please today the teachers are grading it help!!!

Lemur [1.5K]

#6
D. 3.0 x 10^8

#7
A. y = 3.5x + 7

#8
B. 5

3 0

4 years ago

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