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Lynna [10]
2 years ago
6

Please help me quickly

Mathematics
2 answers:
Mademuasel [1]2 years ago
7 0
Give him Brainly 2 people need to answer for Brainly here you go
fiasKO [112]2 years ago
3 0
The answer is B, when you put it into a graphing calculator the Y And X intercept are -5 like in the image
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I giveeeeeeeeeeeeeeeeeeeeeeeeeeee brainlilsttttt
svetlana [45]

Answer:

H

Step-by-step explanation:

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Help with this algebra problem? 12y+6=6y+12?
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First, combine like terms.
12y+6=6y+12 turns into
12y-6y=12-6, which is
6y=6. Now divide both sides by 6.
y=1

now do a quick check (cause that's always a good idea).
12(1)+6=6(1)+12
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3 years ago
What is 1/3 multiply 2/4
Greeley [361]

Answer:

1/3 * 2/4 = 1/3 * 1/2 = (1*1)/(3*2) = 1/6

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3 years ago
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Omar wants to open an account for his grandchildren that he hopes will have
olga_2 [115]

Using compound interest, it is found that he must deposit $56,389.

Compound interest:

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

  • A(t) is the amount of money after t years.  
  • P is the principal(the initial sum of money).  
  • r is the interest rate(as a decimal value).  
  • n is the number of times that interest is compounded per year.  
  • t is the time in years for which the money is invested or borrowed.

In this problem:

  • Hopes to have $80,000 in 20 years, thus t = 20, A(t) = 80000.
  • Interest rate of 1.75%, thus r = 0.0175.
  • Compounding monthly, thus n = 12
  • The investment is of P, for which we have to solve.

Then:

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

80000 = P(1 + \frac{0.0175}{12})^{12(20)}

P = \frac{80000}{(1 + \frac{0.0175}{12})^{12(20)}}

P = 56389

He must deposit $56,389.

A similar problem is given at brainly.com/question/25263233

4 0
3 years ago
Can someone watch number 10 ?
Lerok [7]

Answer:

-9

Step-by-step explanation:

Hello! So, we need to find log_6 1/6. The definition of log is the power that you need to multiply that number by to get the next number. 1/6=6^-1, so our equation would be:

-1=x/9, so x is clearly -9.

Have a nice day.

~cloud

3 0
3 years ago
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