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OlgaM077 [116]
3 years ago
14

Brad wants to open a new savings account. He finds one that returns 3% on his investment each year.

Mathematics
1 answer:
Rufina [12.5K]3 years ago
5 0

Answer:

  • final balance: A
  • initial investment: P
  • return rate: r

Step-by-step explanation:

The variable assigments can be anything you like. It is often helpful to use variables that will remind you what they stand for. Here, we have chosen A = amount (final balance); P = principal (initial investment); r = rate.

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Solve for x. any help would be appreciated thanks
WARRIOR [948]
The correct answer is x=5

8 0
3 years ago
Read 2 more answers
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
I can’t solve this. Please explain!
babunello [35]
Sorry I can’t help....
6 0
3 years ago
Read 2 more answers
Which answer is correct?<br> Thank you
stiv31 [10]

Answer:

Vertical angles

Step-by-step explanation:

: If two angles are vertical angles, then they're congruent (see the above figure). Vertical angles are one of the most frequently used things in proofs and other types of geometry problems, and they're one of the easiest things to spot in a diagram. Don't neglect to check for them!

4 0
2 years ago
Read 2 more answers
Find the general solution of the given differential equation. cos^2(x)sin(x)dy/dx+(cos^3(x))y=1 g
eimsori [14]

If the given differential equation is

\cos^2(x) \sin(x) \dfrac{dy}{dx} + \cos^3(x) y = 1

then multiply both sides by \frac1{\cos^2(x)} :

\sin(x) \dfrac{dy}{dx} + \cos(x) y = \sec^2(x)

The left side is the derivative of a product,

\dfrac{d}{dx}\left[\sin(x)y\right] = \sec^2(x)

Integrate both sides with respect to x, recalling that \frac{d}{dx}\tan(x) = \sec^2(x) :

\displaystyle \int \frac{d}{dx}\left[\sin(x)y\right] \, dx = \int \sec^2(x) \, dx

\sin(x) y = \tan(x) + C

Solve for y :

\boxed{y = \sec(x) + C \csc(x)}which follows from [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}.

7 0
2 years ago
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