Answer:
$2647.13
$2648.08
Step-by-step explanation:
To solve for the value of each loan we will use the formula:
![A=P(1+rt)](https://tex.z-dn.net/?f=A%3DP%281%2Brt%29)
Let's break down the variables that we have.
P = $2,600
r = 7.25% or 0.0725
r2 = 7.50% or 0.0750
t = 90 days
Now since we're computing for two different types of interest, let's take it one at a time.
First the State Saving and Loan.
In this situation we are solving for ordinary interest, where we compute with the total number of days are 360
![A=P(1+rt)](https://tex.z-dn.net/?f=A%3DP%281%2Brt%29)
![A=2,600(1+(0.0725)(\dfrac{90}{360})](https://tex.z-dn.net/?f=A%3D2%2C600%281%2B%280.0725%29%28%5Cdfrac%7B90%7D%7B360%7D%29)
![A=2,600(1+(0.0725)(0.25)](https://tex.z-dn.net/?f=A%3D2%2C600%281%2B%280.0725%29%280.25%29)
![A=2,600(1+0.018125)](https://tex.z-dn.net/?f=A%3D2%2C600%281%2B0.018125%29)
![A=2,600(1.018125)](https://tex.z-dn.net/?f=A%3D2%2C600%281.018125%29)
![A=2,647.13](https://tex.z-dn.net/?f=A%3D2%2C647.13)
The maturity value of State Savings and Loan is $2,647.13.
Now let's move on to the Security bank.
The security bank charges 7.5% exact interest. For exact interest we use 365 days.
![A=2,600(1+(0.0750)(\dfrac{90}{365})](https://tex.z-dn.net/?f=A%3D2%2C600%281%2B%280.0750%29%28%5Cdfrac%7B90%7D%7B365%7D%29)
![A=2,600(1+(0.0750)(0.246575)](https://tex.z-dn.net/?f=A%3D2%2C600%281%2B%280.0750%29%280.246575%29)
![A=2,600(1+(0.0184931)](https://tex.z-dn.net/?f=A%3D2%2C600%281%2B%280.0184931%29)
![A=2,600(1.0184931)](https://tex.z-dn.net/?f=A%3D2%2C600%281.0184931%29)
![A=2,648.08](https://tex.z-dn.net/?f=A%3D2%2C648.08)
The maturity value of the Security bank is $2,648.08.
Answer:
-2x^3 - 3x^2 -8x + 4 should be added
Step-by-step explanation:
Here, we want to know what should be added to the expression to get 0
What we should do in this case is to simply negate the values we have in the expression already
Adding the negated values will give 0
So, we have that;
-2x^3 - 3x^2 -8x + 4
Check the picture below.
first off let's check what "h" is in the triangular face.
![\textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2 \qquad \begin{cases} c=\stackrel{hypotenuse}{10}\\ a=\stackrel{adjacent}{6}\\ b=\stackrel{opposite}{h}\\ \end{cases}\implies 10^2=6^2+h^2 \\\\\\ 10^2-6^2=h^2\implies \sqrt{10^2-6^2}=h\implies 8=h](https://tex.z-dn.net/?f=%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3D%5Cstackrel%7Bhypotenuse%7D%7B10%7D%5C%5C%20a%3D%5Cstackrel%7Badjacent%7D%7B6%7D%5C%5C%20b%3D%5Cstackrel%7Bopposite%7D%7Bh%7D%5C%5C%20%5Cend%7Bcases%7D%5Cimplies%2010%5E2%3D6%5E2%2Bh%5E2%20%5C%5C%5C%5C%5C%5C%2010%5E2-6%5E2%3Dh%5E2%5Cimplies%20%5Csqrt%7B10%5E2-6%5E2%7D%3Dh%5Cimplies%208%3Dh)
so, the gazebo has 4 squarish faces, each 12x12, recall 48 ÷ 4 = 12, and it has 4 triangular faces, each with a height of 8 and a base of 12.
notice we're skipping the top and bottom of the cube and the bottom of the pyramid because they're not part of the "surface area".
![\stackrel{\textit{\large Areas}}{\stackrel{\textit{4 triangular faces}}{4\left[\cfrac{1}{2}(12)(8) \right]}~~~~+~~~~\stackrel{\textit{4 squarish faces}}{4[12\cdot 12]}}\implies 192+576\implies 768](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5Clarge%20Areas%7D%7D%7B%5Cstackrel%7B%5Ctextit%7B4%20triangular%20faces%7D%7D%7B4%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%2812%29%288%29%20%5Cright%5D%7D~~~~%2B~~~~%5Cstackrel%7B%5Ctextit%7B4%20squarish%20faces%7D%7D%7B4%5B12%5Ccdot%2012%5D%7D%7D%5Cimplies%20192%2B576%5Cimplies%20768)
Answer:
20m+60
Step-by-step explanation:
given,
mark=m
june=m+6(6 years older that mark)
so,
after 10 years,
10(m+(m+6)
=10(2m+6)
=20m+60
Answer:
Step-by-step explanation:
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