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likoan [24]
3 years ago
5

Find the value of x.

Mathematics
1 answer:
grandymaker [24]3 years ago
7 0
First do 70+24=94
Then do 180-94=86
Next 86+54=140
Then 180-140=40
Next 40+80=120
Lastly 180-120=60

X=60
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Which equation represents an exponential function with an initial value of 500?
Georgia [21]

Answer: C

Step-by-step explanation:

8 0
3 years ago
Let P(x) denote the statement "2x+5 > 10." Which of the following is true?
ruslelena [56]

Answer: P(3) is True

Step-by-step explanation:

The given statement is an inequality denoted as P(x). To find out which of the options is true you have to evaluate each given value of X in the inequality and perform the arithmetic operations, then you have to see if the expression makes sense.

For P(0): Replace X=0 in 2x+5>10

2(0)+5>10

0+5>10

5>10 is false because 5 is not greater than 10

For P(3): Replace X=3 in 2x+5>10

2(3)+5>10

6+5>10

11>10 is true because 11 is greater than 10

For P(2): Replace X=2 in 2x+5>10

2(2)+5>10

4+5>10

9>10 is false

For P(1): Replace X=1 in 2x+5>10

2(1)+5>10

2+5>10

7>10 is false

4 0
3 years ago
HELPP please.. please pleaae
Zinaida [17]

Answer:

A=80

B=31

C=64

Step-by-step explanation:

A, 69, and b will equal 180 since it forms a straight line

A+69+b=180

to solve for A we subtract 53+47 from 180 and get 80

80+69+b=180

to get B we subtract 80+69 from 180 and get 31

to get c we add 85+b and subtract it from 180 getting c=64

6 0
3 years ago
Read 2 more answers
Are these correct? If not please help me write the correct answer.​
Keith_Richards [23]

Answer:

Mostly correct

For Q3, you are doing great, you remembered that negative numbers can happen if the exponent is an odd number.

For Q4, I think you were confused with the fractions,

4(i)

(\frac{2}{3} )^{4} = (\frac{2^{4}}{3^{4}}) = (\frac{16}{81}) =0.198

4(ii) is also wrong but I'll let you try to fix it yourself following what I given you for 4(i). You should get 1/64

Hope this helps!

4 0
3 years ago
Read 2 more answers
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
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