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evablogger [386]
2 years ago
11

[BRAINLIEST] a. Nick deposits money in a Certificate of Deposit account (CD). The balance (in dollars) in his account t years af

ter making the deposit is given by N(t)=400(1.06)tfor t ≥ 0. Explain, in terms of the structure of the expression for N(t), why Nick's balance can never be $399.
Mathematics
1 answer:
crimeas [40]2 years ago
4 0

Answer:

(1.06)0 = 1  and positive powers of 1.06 are larger than 1, thus the minimum value N(t) attains, if t≥0, is 400.

From the point of view of the context, a CD account grows in value over time so with a deposit of $400 the value will never drop to $399.

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Is it essary to rename 4 1/4 if you subtract 3/4 from it.
Archy [21]
No . because you would then have 2/4 which is 1/2 50%

3 0
3 years ago
A geometric sequence is defined by the general term tn = 75(5n), where n ∈N and n ≥ 1. What is the recursive formula of the sequ
andreyandreev [35.5K]
The correct answer is C) t₁ = 375, t_n=5t_{n-1}.

From the general form,
t_n=75(5)^n, we must work backward to find t₁.

The general form is derived from the explicit form, which is
t_n=t_1(r)^{n-1}.  We can see that r = 5; 5 has the exponent, so that is what is multiplied by every time. This gives us

t_n=t_1(5)^{n-1}

Using the products of exponents, we can "split up" the exponent:
t_n=t_1(5)^n(5)^{-1}

We know that 5⁻¹ = 1/5, so this gives us
t_n=t_1(\frac{1}{5})(5)^n
\\
\\=\frac{t_1}{5}(5)^n

Comparing this to our general form, we see that
\frac{t_1}{5}=75

Multiplying by 5 on both sides, we get that
t₁ = 75*5 = 375

The recursive formula for a geometric sequence is given by
t_n=t_{n-1}(r), while we must state what t₁ is; this gives us

t_1=375; t_n=t_{n-1}(5)

3 0
3 years ago
Solve r=4/3(p-q) for p
lesya [120]

Answer:

q+3/4r=p

Step-by-step explanation:

r=4/3(p-q)

Distribute the 4/3

r=4/3p-4/3q

Add 4/3q to each side

4/3q+r=4/3p

Multiply ALL variables by 3/4 (undoes the 4/3)

q+3/4r=p

<h2>Please mark me as brainliest</h2>

8 0
2 years ago
PLEASE SHOW WORK AND SIMPLIFY
BabaBlast [244]

Answer:

Step-by-step explanation:

5 0
3 years ago
Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent
goblinko [34]

Answer:

y(x)=x^2+5x

Step-by-step explanation:

Given: y'=\sqrt{4y+25}

Initial value: y(1)=6

Let y'=\dfrac{dy}{dx}

\dfrac{dy}{dx}=\sqrt{4y+25}

Variable separable

\dfrac{dy}{\sqrt{4y+25}}=dx

Integrate both sides

\int \dfrac{dy}{\sqrt{4y+25}}=\int dx

\sqrt{4y+25}=2x+C

Initial condition, y(1)=6

\sqrt{4\cdot 6+25}=2\cdot 1+C

C=5

Put C into equation

Solution:

\sqrt{4y+25}=2x+5

or

4y+25=(2x+5)^2

y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}

y(x)=x^2+5x

Hence, The solution is y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4} or y(x)=x^2+5x

4 0
3 years ago
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