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svetoff [14.1K]
2 years ago
5

Find the measure of BCD

Mathematics
1 answer:
Lerok [7]2 years ago
4 0

<em><u>p</u></em><em><u>l</u></em><em><u>e</u></em><em><u>a</u></em><em><u>s</u></em><em><u>e</u></em><em><u> </u></em><em><u>m</u></em><em><u>a</u></em><em><u>r</u></em><em><u>k</u></em><em><u> </u></em><em><u>m</u></em><em><u>e</u></em><em><u> </u></em><em><u>a</u></em><em><u>s</u></em><em><u> </u></em><em><u>b</u></em><em><u>r</u></em><em><u>a</u></em><em><u>i</u></em><em><u>n</u></em><em><u>l</u></em><em><u>i</u></em><em><u>e</u></em><em><u>s</u></em><em><u>t</u></em><em><u>!</u></em><em><u>!</u></em>

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Solve the equation using square roots.<br><br> 2x^2-98=0
Yuliya22 [10]

Step-by-step explanation:

2x²-98=0

2x²=98

x²=98/2

x²=49

x=root 49

so,x=7

if it is correct answer then please follow me

3 0
3 years ago
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}

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2 years ago
Radius of a sphere that has a surface area 12.56in2 use 3.14
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8 0
3 years ago
What is m in this question
solmaris [256]
-16 because you need to subtract -4 from -12
7 0
3 years ago
I have a question?.....Ms. Wallace makes soap to sell at craft fairs. The first bar of soap sells for $2, and each additional ba
Assoli18 [71]

Answer:

Has your teacher given you notes to practice?

Step-by-step explanation:

If no, then break the question down into pieces and write down what you know.

3 0
2 years ago
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