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kari74 [83]
3 years ago
5

Rod is saving Php2000 in a bank at the end of each month which gives an interest of 1% compounded monthly.How much is the saving

s og Rod after 2 years? ​
Mathematics
1 answer:
agasfer [191]3 years ago
7 0

Answer:

Php2040.38

Step-by-step explanation:

Given

P = 2000 --- Principal

r = 1\% --- Rate

t = 2\ years --- Time

n = 12 --- monthly

Required

Determine the amount at the end of two years

This is calculated as:

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

So, we have:

A = 2000(1 + \frac{1\%}{12})^{12*2}

A = 2000(1 + \frac{1}{100*12})^{24}

A = 2000(1 + \frac{1}{1200})^{24}

A = 2000(\frac{1200+1}{1200})^{24}

A = 2000(\frac{1201}{1200})^{24}

A = 2000*1.02019

A = 2040.38

<em>Hence, the final amount is: Php2040.38</em>

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5 0
3 years ago
Order the side lengths from least to greatest
Liula [17]

Answer:

BC < CE < BE < ED < BD

Step-by-step explanation:

In the triangle BCE,

m∠BEC + m∠BCE + m∠CBE = 180°

m∠BEC + 81° + 54° = 180°

m∠BEC = 180 -  135

m∠BEC = 45°

Order of the angles from least to greatest,

m∠BEC < m∠CBE > mBCE

Sides opposite to these sides will be in the same ratio,

BC < CE < BE ----------(1)

Now in ΔBED,

m∠BEC + m∠BED = 180°

m∠BED = 180 - 45

             = 135°

Now, m∠BDE + m∠BED + DBE = 180°

11° + 135°+ m∠DBE = 180°

m∠DBE = 180 - 146

             = 34°

Order of the angles from least to greatest will be,

∠BDE < ∠DBE < ∠BED

Sides opposite to these angles will be in the same order.

BE < ED < BD ----------(2)

From relation (1) and (2),

BC < CE < BE < ED < BD

7 0
3 years ago
What is the square root of 16x^<br> 36?
Vsevolod [243]

Note that \sqrt[2]{x}=x^{\frac{1}{2}}

\sqrt[2]{16x^{36}}=\sqrt[2]{4^2x^{18\cdot2}}

4^{2\cdot\frac{1}{2}}x^{18\cdot2\cdot\frac{1}{2}}=4^{\frac{2}{2}}x^{\frac{18\cdot2}{2}}

\boxed{4x^{18}}

Hope this helps.

r3t40

4 0
3 years ago
Answer asap thank you!??!!??!
Semenov [28]
The answer to the problem is A
7 0
3 years ago
Find the standard form of the circle x2 + y2 – 6x + 10y + 24 = 0 by completing the square.
hichkok12 [17]

Answer:

(x-3)^2+ (y+5)^2=10

Step-by-step explanation:

Given the equation of the circle: x^2 + y^2 - 6x + 10y + 24 = 0

We wish to express it in a Standard form:

We begin by re-arranging:

x^2 - 6x + y^2  + 10y  = - 24

Next, divide the coefficient of x by 2, square it and add it to both sides.

Do the same for y.

x^2 - 6x +(-3)^2+ y^2  + 10y+5^2  = - 24+(-3)^2+5^2

Next, we factorize

(x-3)^2+ (y+5)^2  = - 24+9+25\\(x-3)^2+ (y+5)^2=10

This is the standard form.

6 0
3 years ago
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