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guajiro [1.7K]
3 years ago
15

What is the simplified form of the following expression?

Mathematics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

242

Step-by-step explanation:

Simplify the following:

11 ((9^2 - 5^2)/2^2 + 8)

Hint: | Evaluate 2^2.

2^2 = 4:

11 ((9^2 - 5^2)/4 + 8)

Hint: | Evaluate 5^2.

5^2 = 25:

11 ((9^2 - 25)/4 + 8)

Hint: | Evaluate 9^2.

9^2 = 81:

11 ((81 - 25)/4 + 8)

Hint: | Subtract 25 from 81.

| 7 | 11

| 8 | 1

- | 2 | 5

| 5 | 6:

11 (56/4 + 8)

Hint: | Reduce 56/4 to lowest terms. Start by finding the GCD of 56 and 4.

The gcd of 56 and 4 is 4, so 56/4 = (4×14)/(4×1) = 4/4×14 = 14:

11 (14 + 8)

Hint: | Evaluate 14 + 8 using long addition.

| 1 |  

| 1 | 4

+ | | 8

| 2 | 2:

11×22

Hint: | Multiply 11 and 22 together.

| 2 | 2

× | 1 | 1

| 2 | 2

2 | 2 | 0

2 | 4 | 2:

Answer: 242

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astra-53 [7]

Answer:

$110.70

Step-by-step explanation:

<em>I=PRT</em>

P (principal)=360

R(rate)=12.3%

T(time)=30 months (2 and half years)

I=360*12.3%*2.5

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Just to note 12.3% is also 0.123 we sometimes change it into a decimal.

3 0
2 years ago
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April shoots an arrow upward at a speed of 80 ft/sec from a platform of 25 ft. High. The pathway of the arrow can be represented
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Answer:

125feet

Step-by-step explanation:

Given the equation that modeled the height expressed as h = -16t^2 + 80t + 25, where h is the height and t is the time in seconds.

The arrow reaches the maximum height at dh/dt = 0

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0= -32t+80

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t = 80/32

t = 2.5secs

substitute t = 2.5 into the formula;

h = -16t^2 + 80t + 25

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h = -16(6.25)+225

h = -100+225

h = 125

Hence the maximum height the arrow reaches is 125feet

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At which value in the domain does f(x) = 0?<br> 2<br> 2<br> 4
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4

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer with  explanation:

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