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antiseptic1488 [7]
3 years ago
10

What is the value of 6x – 3y if x = 5 and y=-1?F. 11 G. 33 H. 65 1. 65​

Mathematics
2 answers:
atroni [7]3 years ago
7 0

Answer:

it is G.33

brainliest plz

Step-by-step explanation:

SOVA2 [1]3 years ago
7 0

Answer: 33

Step-by-step explanation:

6(5)-3(-1)

30+3

33

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Thang decided to borrow $2000 from his local bank to help pay for a car. His loan was for 3 years at a simple intrest rat it 5%.
irakobra [83]

Answer:Thang will pay $300 as interest

Step-by-step explanation:

We would apply the formula for determining simple interest which is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time in years

I = interest after t years

From the information given

T = 3 years

P = $2000

R = 5%

Therefore

I = (2000 × 5 × 3)/100

I = 30000/100

I = $300

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3 years ago
In a linear Pair angles are?
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2 years ago
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B..............
c>=3 and n<0 and p<=(9-3 c)/n
8 0
3 years ago
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How do you solve 25x = (1/125)^(4x-5)
amm1812

Step-by-step explanation:

Step 1: 25*x-((1/125)^(4*x-5))=0

step 2: simplify 1/125

step 3: 25x-(1/125)(4x-5)=0

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2 years ago
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Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete
Katyanochek1 [597]

Answer:

  1. b/a
  2. 16a²b²
  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

__

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

__

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

__

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

__

4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

__

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}

3 0
3 years ago
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