8) Solve: X + 250 = 700

Find the number of 4 digit numbers that can be formed using the digits 1,2,3,4,5 if no digits is no repeated

loris [4]

Answer:

5 different numbers, 2 even

Step-by-step explanation:

Since we have 5 digits and they are non repeating we can use the permutation formula:

$P(n, k) = \frac{n!}{k!}, $where $ k \leq n ; k, n \in \mathbb{N}$

It gives us the amount of ways to arrange k objects out of n objects without repetition, but order does matter.

In our case the amount of 4 digit numbers that can be formed using the digits from 1 to 5 is: $P(5, 4) = \frac{5!}{4!} = 5.$

And out of those 5, only 2 are even

4 0

2 years ago

Sven got a $150 gift card function A(n) 150 12.30n gives the amount of money left on the gift card after n number of visits to t

lorasvet [3.4K]

Answers:

1. The n-intercept is 12. That means after 12 visits the amount of money on the gift car is $0.

2. The A(n)-intercept is 150. Before the visits, the amount of money on the gift car is $150.

Solution:

Amount of money on the gift card after n number of visits: A(n)=$150-$12.50 n

A(n)=150-12.50 n

1. n-intercept

A(n)=0→150-12.50 n =0

Solving for n: Subtracting 150 both sides of the equation:

150-12.50 n-150 = 0-150

-12.50 n = -150

Dividing both sides of the equation by -12.50:

(-12.50 n) / (-12.50) = (-150) / (-12.50)

n=12

The n-intercept is n=12; for n=12→A(12)=0. Point (n, A(n))=(12,0)

2. A(n) intercept

n=0→A(0)=150-12.50 (0)

A(0)=150-0

A(0)=150

The A(n) intercept is 150; for n=0→A(0)=150. Point (n, A(n))=(0,150)

8 0

2 years ago

Evaluate the surface integral ∫sf⋅ ds where f=⟨2x,−3z,3y⟩ and s is the part of the sphere x2 y2 z2=16 in the first octant, with

skad [1K]

Parameterize S by the vector function

$\vec s(u,v) = \left\langle 4 \cos(u) \sin(v), 4 \sin(u) \sin(v), 4 \cos(v) \right\rangle$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2.

Compute the outward-pointing normal vector to S :

$\vec n = \dfrac{\partial\vec s}{\partial v} \times \dfrac{\partial \vec s}{\partial u} = \left\langle 16 \cos(u) \sin^2(v), 16 \sin(u) \sin^2(v), 16 \cos(v) \sin(v) \right\rangle$

The integral of the field over S is then

$\displaystyle \iint_S \vec f \cdot d\vec s = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \vec f(\vec s) \cdot \vec n \, du \, dv$

$\displaystyle = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \left\langle 8 \cos(u) \sin(v), -12 \cos(v), 12 \sin(u) \sin(v) \right\rangle \cdot \vec n \, du \, dv$

$\displaystyle = 128 \int_0^{\frac\pi2} \int_0^{\frac\pi2} \cos^2(u) \sin^3(v) \, du \, dv = \boxed{\frac{64\pi}3}$

8 0

1 year ago

What is a million times all the numbers in the world

kozerog [31]

Well, there isn’t really an end for numbers...

However; The biggest number referred to regularly is a googolplex (10googol), which works out as 1010^100. That isn’t the end to numbers but it is a huge one. We will replace that with ‘all the numbers in the world’.

106 is the exponent equivalent to 1 million

So your question would be:
106 x 1010^100 =

However I don’t believe there is a calculator that large.

7 0

2 years ago

Could someone help and explain their answer? (NO LINKS!)<br> Thanks!

Nonamiya [84]

Answer:

5

Step-by-step explanation:

To find the prime factorization of a number, start by dividing the number by each prime number that you can divide by as many times as possible, starting with the smallest prime number and moving up. Then move on to the next prime number. The last division must give a quotient of 1. Then the prime factorization is the product of all the prime factors you divided the number by.

Here's how you use this method with 180.

Here are the first 10 prime numbers:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Start by trying to divide by 2. 180 is divisible by 2.

180/2 = 90 90 is divisible by 2.

90/2 = 45 45 is not divisible by 2, so now try 3.

45/3 = 15 15 is divisible by 3.

15/3 = 5 5 is prime, so it is divisible by 5.

5/5 = 1 The divisors in bold are the prime factors of 180.

180 = 2 × 2 × 3 × 3 × 5

180 = 2² × 3² × 5

Answer: 5

3 0

2 years ago

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