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makvit [3.9K]
3 years ago
15

I honestly don’t understand this ::))

Mathematics
1 answer:
GrogVix [38]3 years ago
7 0

Answer:

Answer is 100 sq m

Step-by-step explanation:

A = a^{2}

A = 10^{2}

A = 100 sq m

HOPE IT HELPS

<u><em>(FROM CROSS)</em></u>

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If F (x) =8+11 X-3X², find<br> a. What is F (5)? <br> b. What is F (x + b)? <br> c. What is F (-3)?
stepladder [879]

Answer:

a) -12

b) 8+11x+11b-3x^2-3b^2-6xb

c)-52

Step-by-step explanation:

in a) we have to substitute the value of x with 5

in b) we substitute the value of x with (x+b) and then simplify

in c) just have to substitute x with -3

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Help me- what number of units and where
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3 years ago
What’s the value of x? :/
kolbaska11 [484]

Answer:

x = 27

Step-by-step explanation:

According to congruent chords and arcs theorem, if \overline{XY} \cong \overline{ZY} then \arc{XY} \cong \arc{ZY}.

This means that \arc{XY} = \arc{ZY} = (6x - 20) degrees

Thus:

\arc{XY} + \arc{ZY} + \arc{ZX} = 360 (full circle = 360°)

\arc{XY} = (6x - 20)

\arc{ZY} = (6x - 20)

\arc{ZX} = 76

Plug in the values into the equation

(6x - 20) + (6x - 20) + 76 = 360

6x - 20 + 6x - 20 + 76 = 360

Add like terms

12x + 36 = 360

Subtract 36 on both sides

12x = 360 - 36

12x = 324

Divide both sides by 12

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8 0
3 years ago
(Brainliest) PLEASE HELP PLEASE
lesantik [10]

what's the question?

4 0
3 years ago
Read 2 more answers
Occasionally a savings account may actually pay interest compounded continuously. For each deposit, find the interest earned if
Ugo [173]

1. Occasionally a savings account may actually pay interest compounded continuously. For each​ deposit, find the interest earned if interest is compounded​ (a) semiannually,​ (b) quarterly,​ (c) monthly,​ (d) daily, and​ (e) continuously. Use 1 year = 365 days.

Principal ​$1031

Rate 1.4%

Time 3 years

Answer:

a) $ 44.07

b) $ 44.15

c) $ 44.20

d) $ 44.22

e) $ 44.22

Step-by-step explanation:

The formula to find the total amount earned using compound interest is given as:

A = P(1 + r/n)^nt

Where A = Total amount earned after time t

P = Principal = $1031

r = Interest rate = 1.4%

n = compounding frequency

t = Time in years = 3 years

For each​ deposit, find the interest earned if interest is compounded

(a) semiannually

This means the interest is compounded 2 times in a year

Hence:

A = P(1 + r/n)^nt

A = 1031(1 + 0.014/2) ^2 × 3

A = 1031 (1 + 0.007)^6

A = $ 1,075.07

A = P + I where

I = A - P

I = $1075.07 - $1031

P (principal) = $ 1,031.00

I (interest) = $ 44.07

​(b) quarterly

This means the interest is compounded 4 times in a year

Hence:

A = P(1 + r/n)^nt

A = 1031(1 + 0.014/4) ^4 × 3

A = 1031 (1 + 0.014/4)^12

A = $ 1,075.15

I = A - P

I = $1075.15 - $1031

A = P + I where

P (principal) = $ 1,031.00

I (interest) = $ 44.15

(c) monthly,

​ This means the interest is compounded 12 times in a year

Hence:

A = P(1 + r/n)^nt

A = 1031(1 + 0.014/12) ^12 × 3

A = 1031 (1 + 0.014/12)^36

A = $ 1,075.20

A = P + I where

I = A - P

I = $1075.20 - $1031

P (principal) = $ 1,031.00

I (interest) = $ 44.20

(d) daily,Use 1 year = 365 days

This means the interest is compounded 365 times in a year

Hence:

A = P(1 + r/n)^nt

A = 1031(1 + 0.014/365) ^2 × 3

A = 1031 (1 + 0.00365)^365 × 3

A = $ 1,075.22

A = P + I where

I = A - P

I = $1075.22 - $1031

P (principal) = $ 1,031.00

I (interest) = $ 44.22

(e) continuously. .

This means the interest is compounded 2 times in a year

Hence:

A = Pe^rt

A = 1031 × e ^0.014 × 3

A = $ 1,075.22

A = P + I where

I = A - P

I = $1075.22 - $1031

P (principal) = $ 1,031.00

I (interest) = $ 44.22

5 0
3 years ago
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