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GarryVolchara [31]
3 years ago
7

Use a factor tree to find the prime factorizations of 96. Write the prime factorizations using exponents. 1.2^3•6^2. 2.2^5•4^2.3

. 2^5•3. 4.3•5^2

Mathematics
1 answer:
dezoksy [38]3 years ago
4 0
Look\ at\ the\ picture:\\\\96=2\times2\times2\times2\times2\times3=\boxed{2^5\cdot3}\leftarrow\ answer\ \boxed{3.\ 2^5\cdot3}

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Use the Pythagorean identity to do the following:
Kay [80]

Answer: 2sin(θ)

Step-by-step explanation:       (1 − cos2(theta)) csc(theta)  

cos (2θ) = cos²(θ) - sin²(θ)     ⇒ [(1 - cos²(θ) + sin²(θ)]* csc (θ)

Multiplying        csc(θ) - cos²(θ)*csc(θ) + sin²(θ)*csc(θ)

csc (θ) = 1 ÷ sin (θ)         ⇒   1/sin(θ) - cos²(θ)/sin(θ) + sin²(θ)/sin(θ)

1/sin(θ) [ 1 -cos²(θ) + sin²(θ)]    ⇒  1- cos²(θ) = sin²θ)

1/sin(θ) [ sin²(θ) +sin²(θ) ]

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7 0
3 years ago
Add the two expressions.<br><br> 6q + 1 and q + 11
Natali [406]
The answer to this equation is: 7q +12
3 0
4 years ago
Please help me please !!
sweet [91]

Answer:

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Step-by-step explanation:

7² + 9² ? 11²

49 + 81 ? 121

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8 0
3 years ago
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Step-by-step explanation:

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8 0
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Read 2 more answers
Write a function to represent each problem situation Carmen deposits $1000into simple interest account.The rate for the account
kolezko [41]

The function the represent the balance in the account as a function of time t is p(t) = 1000 + 40t

<h3><u>Solution:</u></h3>

Given that,

Carmen deposits $1000 into simple interest account

The rate for the account is 4%

To find: function the represent the balance in the account as a function of time t

Given is simple interest account

The formula for simple interest is given as:

S.I = \frac{ prt}{100}

Where, "p" is the principal and "r" is the rate of interest and "t" is the number of years

In simple interest,

total amount after "t" years = principal + simple interest

Here in this question, Carmen deposits $1000

p_0 = 1000

r = 4 \% = \frac{4}{100} = 0.04

Thus we can frame a function as:

total amount after "t" years = principal + simple interest

p(t) = p_0 + (p_0 \times r \times t )

p(t) = 1000 + (1000 \times 0.04)t\\\\p(t) = 1000 + 40t

Where, p(t) is the amount after "t" years and p_0 is the principal sum

Thus the function is obtained

7 0
4 years ago
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