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sineoko [7]
3 years ago
5

If you put $750 into a savings account that earns 6.2%, how much interest will you receive at the end of 7 years?

Mathematics
1 answer:
frozen [14]3 years ago
5 0

Answer: You will receive $ 325.5 as interest at the end of 7 years.

Step-by-step explanation:

Given: Principal amount = $750

Rate of interest = 6.2% = 0.062

Time = 7 years

Interest = Principal x Rate x Time

Inetrest = 750 x 0.062 x 7

= $ 325.5

Hence, you will receive $ 325.5 as interest at the end of 7 years.

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(x+3)²(65)=x(6)<br>Solve using quadric formula​
Angelina_Jolie [31]

Step 1: Simplify both sides of the equation.

  • 65x²+390x+585=6x

Step 2: Subtract 6x from both sides.

  • 65x²+390x+585−6x=6x−6x
  • 65x²+384x+585=0

For this equation: a=65, b=384, c=585

  • 65x²+384x+585=0

Step 3: Use quadratic formula with a=65, b=384, c=585

  • x=−b±√b2−4ac/2a
  • x=−(384)±√(384)2−4(65)(585)/
  • 2(65)
  • x=−384±√−4644/130

Therefore, There are no real solutions.

6 0
3 years ago
Read 2 more answers
Plz gyz ineed the calculation
postnew [5]

Answer:

1/8

Step-by-step explanation:

1st day......1/3

2nd day....1/6

3rd day....3/8

common denominator of 24

1/3 + 1/6 + 3/8 = 8/24 + 4/24 + 9/24 = 21/24 (this is water used)

that means that : (24/24 - 21/24) = 3/24 reduces to 1/8 .....water in the tank

6 0
4 years ago
Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions
monitta
I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take z=y', so that z'=y'' and we're left with the ODE linear in z:

y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2

Now suppose y has a power series expansion

y=\displaystyle\sum_{n\ge0}a_nx^n
\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Then the ODE can be written as

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0

\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0

All the coefficients of the series vanish, and setting x=0 in the power series forms for y and y' tell us that y(0)=a_0 and y'(0)=a_1, so we get the recurrence

\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}

We can solve explicitly for a_n quite easily:

a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}

and so on. Continuing in this way we end up with

a_n=\dfrac{a_1}{n!}

so that the solution to the ODE is

y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x

We also require the solution to satisfy y(0)=a_0, which we can do easily by adding and subtracting a constant as needed:

y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}
4 0
3 years ago
Please help me! please!
netineya [11]

Answer:

(c)

Step-by-step explanation:

(c). Stopping for traffic

5 0
3 years ago
I will give brainliest if correct
dalvyx [7]

Answer:supplementary

Step-by-step explanation:

Opposite angles in any quadrilateral inscribed in a circle are supplements of each other.

5 0
3 years ago
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