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dalvyx [7]
1 year ago
8

Rajesh invested $30,000 into an account that compounds interest monthly at a rate of 2.16%. He has made arrangements to withdraw

$300 automatically every month to pay off his 10-year student loan. Will Rajesh have enough money in the account to cover all of the required loan payments? (Round to the nearest tenth of a year.)
Mathematics
1 answer:
Bond [772]1 year ago
5 0

By Evaluating the Compound Interest, we come to know that Rajesh will have enough money in the account to cover all of the required loan payments.

The Principal Amount(P) = $30,000

Rate of Interest (r) = 2.16 %

Time(t) = 10 years

Number of Times it is Compounded in a year(n) = 12

Now, we have

A =P(1+\frac{r}{100n}) ^{nt}

Putting all the values, we evaluate the amount,

A =30,000(1+\frac{2.16}{100*12}) ^{12*10}\\\\A = 30,000 * 1.240\\A = 37,225.84

Hence, the Amount after Compound Interest = $37,225.87

Now, The loan amount that he pays = 300 *12*10 = $ 36,000

Yes, he will have enough money in the account to cover all of the required loan payments.

To read more about Compound Interest, visit brainly.com/question/29335425

#SPJ1

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Answer:

80 m^2

Step-by-step explanation:

The formula is lwh/3. 5*6*8=240   240/3=80

7 0
2 years ago
A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la
SOVA2 [1]

Answer:

8\pi\text{ square cm}

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

S=2\pi rh

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

4^2 = (2r)^2 + h^2   ( see in the below diagram ),

16 = 4r^2 + h^2

16 - 4r^2 = h^2

\implies h=\sqrt{16-4r^2}

Thus, the surface area of the cylinder,

S=2\pi r(\sqrt{16-4r^2})

Differentiating with respect to r,

\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})

=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})

=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})

Again differentiating with respect to r,

\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})

For maximum or minimum,

\frac{dS}{dt}=0

2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0

-8r^2 + 16 = 0

8r^2 = 16

r^2 = 2

\implies r = \sqrt{2}

Since, for r = √2,

\frac{d^2S}{dt^2}=negative

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

S = 2\pi (\sqrt{2})(\sqrt{16-8})

=2\pi (\sqrt{2})(\sqrt{8})

=2\pi \sqrt{16}

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4 0
2 years ago
4/5 x = 8 i need to show my work​
Anika [276]

Answer:

x=10

Step-by-step explanation:

I'm assuming you needed to solve for x. Here is how you do it:

first, divide both sides by 4/5 to get x by itself.

this will give you x=8 divided by 4/5.

To solve 8 divided by 4/5 you do keep switch flip.

keep the first number (8), switch the sign (division to multiplication), and flip the other fraction (4/5 to 5/4) then you can just multiply across the top (8/1 *5/4) which equals 40/4. This simplified is 10

4 0
2 years ago
In ΔABC, AD and BE are the angle bisectors of ∠A and ∠B and DE║ AB. Find the measures of the angles of ΔABC, if m∠ADE: m∠ADB = 2
drek231 [11]

Answer:   ∠A=48°,∠B=48°,∠C=84°.


Step by-step explanation:

Given:  AD and BE are the angle bisectors  of ∠A and ∠B

i.e ∠6=∠7    ( ∵ Angles formed after  AD bisected ∠A)

∠4=∠5       ( ∵ Angles formed after  BE bisected ∠B)

Also,  DE║AB

⇒ ∠2=∠7   (∵ Alternate interior angles)

   ∠3=∠6   (∵ Alternate interior angles)

And ∠ADE : ∠ADB =∠2:∠3= 2:9 =2x : 9x     ..(1)

To Find:  ∠A,∠B,∠C.

Solution:  ∠2=∠7 (∵ Given)    ...(2)

∠2=∠4  (∵ angles on the same segment)    ...(3)

∠4=∠5 =∠B/2 (∵ Given)    ...(4)

∴ In Δ ABD

∠3+∠4+∠5+∠7 = 180 (∵ Sum of interior angles of a triangle)

From equation 2,3,4,5, Put values

9x+2x+2x+2x =180°

⇒15x = 180°

⇒x=12°

Putting values in equation (4) ⇒ ∠ B =2*(2*12) = 48°

Also, <u>∠B=∠A=48°</u>

Now,in Δ ABC

∠C+∠B+∠A= 180°

⇒48°+48°+∠C= 180°

<u><em>⇒∠C=84°</em></u>

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Irina-Kira [14]
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