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pickupchik [31]
3 years ago
12

Write in point-slope form.

Mathematics
2 answers:
Mrrafil [7]3 years ago
8 0
The answer is D) y-1=-3(x+2)
hichkok12 [17]3 years ago
7 0

Answer:

the answer is D y-1=-3(x+2)

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Compare. Use <,>, or = to complete the statement
expeople1 [14]
Its >
( {2}^{7} )^{2}  > ( {2}^{10}  {2}^{2})  \\  {2}^{14}  >  {2}^{12}
5 0
3 years ago
Please help!! find x
Marat540 [252]

Answer:

12

Step-by-step explanation:

(25 + x)² = x² + 35²

625 + 50x + x² = x² + 1225

50x = 600

x = 12

3 0
4 years ago
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Mary Hernandez had a policy with a $250 deductible which paid 80% of her covered charges less deductible. She had medical expens
Nataly_w [17]
A. the insurance company's payment = $14,392
$18240 - $250 = 17990 x 80% = $14,392

b. <span>the 20% copayment
17990 * 20% = $3598
</span>
<span>c. Mary's total cost
$250 + $3598 = $3848
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6 0
3 years ago
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The nth term of a sequence is 3n^2- 1
Thepotemich [5.8K]

Answer: The number that belongs to both sequences is 26.

Step-by-step explanation:

We have two sequences, let's call one as A and the other as B.

The n-th term of sequence A is written as:

aₙ = 3*n^2 - 1

the nth term of sequence B is written as:

bₙ = 30 - n^2

We want to find a term that belongs to both sequences, (it can be for different integers, we can use n for sequence A and x for sequence B)

Then we want to find:

aₙ = bₓ

where n and x are integer numbers.

Then we will heave:

3*n^2 - 1 = 30 - x^2

To find the pair, we could isolate one of the variables, then input different integers in the other variable and see if the outcome is also an integer.

Let's isolate n.

3*n^2 = 30 - x^2 + 1

3*n^2 = 31 - x^2

n^2 = (31 - x^2)/3

n = √(  (31 - x^2)/3)

Now let's input different values for x, and see if the outcome is also an integer, notice that x is in a negative term inside a square root, then we have only a few values of x such that the equation can be true.

Then let's start with x = 1.

n(1) = √(  (31 - 1^2)/3) = √(30/3) = √10

We know that √10 is not an integer.

now with x = 2,

n(2) = √(  (31 - 2^2)/3)  = √( (31 - 4)/3) = √(27/3) = √9 = 3

then if x = 2, we have n = 3.

Both of them are integers, then we get:

a₂ = 3*(3)^2 - 1 = 27 - 1 = 26

b₃ = 30 - 2^2 = 30 - 4 = 26

The number that belongs to both sequences is 26.

5 0
3 years ago
Need help , have to summit this soon
ollegr [7]

Answer:

(a). 32 copies

(b). $3060

Step-by-step explanation:

<u>Solution for (a):</u>

<u />

P = 60c + 1800

If P is $3720 then;

3720 = 60c + 1800

60c = 3720 - 1800 = 1920

c = \frac{1920}{60} = 32 copies

<u>Solution for (b):</u>

<u />

P = 60c + 1800

If c is 21 then;

P = $60(21) + $1800

P = $1260 + $1800

P = $3060

8 0
3 years ago
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