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Nataly [62]
2 years ago
11

What is 10/18 equivalent to

Mathematics
1 answer:
a_sh-v [17]2 years ago
7 0

Answer:

5/9

Step-by-step explanation:

in order to find what 10/18 is equivalent to, you have to simplify it. To simplify it you find a number that 10 and 18 are both divisible by,2, and divide each number by it until you get prime numbers or numbers that went divisible by the same number.

I hope this helped.

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BD bisects ∠ABC, m∠ABD=(7x−1)°, and m∠DBC=(4x+8)°. Find m∠ABD.
Naya [18.7K]

Answer:

  m∠ABD = 20°

Step-by-step explanation:

The bisector makes angles ABD and BDC congruent, so ...

  7x -1 = 4x +8

  3x = 9 . . . . . . add 1-4x to both sides

  x = 3

  m∠ABD = (7x-1)° = (7·3 -1)°

  m∠ABD = 20°

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3 years ago
The measures of ∠x and ∠y are equal. Which is the value of m?
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There is no way to tell because there is no value for x and y. However just looking at the picture, I’m guessing m is around 45 degrees.
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3 years ago
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The statement explains why the ordered pair is a solution to the system of equations. Is the statement true or false? True False
tigry1 [53]

The system of the equation doesn't give the solution at (-3, -6).

<h2>Given to us</h2>
  • -4x+y = 6
  • 5x-y =21

<h3>Equation 1,</h3>

-4x+y = 6

solve for y

-4x+y =6\\y= 6+4x

<h3>Equation 2,</h3>

5x-y =21

substitute the value of y in equation 2,

5x-(6+4x) =21\\5x -6 -4x =21\\5x -4x=21+6\\x = 27

Substitute the value of x in equation 2,

5x-y =21\\5(27)-y =21\\135-y = 21\\-y=21-135\\-y = -114\\y=114

We can see that the solution of the two equations is at (27, 114). Also, it can be verified by plotting the line on the graph.

Hence, the system of the equation doesn't give the solution at (-3, -6).

Learn more about system of equations:

brainly.com/question/12895249

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2 years ago
Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference
Bezzdna [24]

Since a_1,a_2,a_3,\cdots are in arithmetic progression,

a_2 = a_1 + 2

a_3 = a_2 + 2 = a_1 + 2\cdot2

a_4 = a_3+2 = a_1+3\cdot2

\cdots \implies a_n = a_1 + 2(n-1)

and since b_1,b_2,b_3,\cdots are in geometric progression,

b_2 = 2b_1

b_3=2b_2 = 2^2 b_1

b_4=2b_3=2^3b_1

\cdots\implies b_n=2^{n-1}b_1

Recall that

\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n

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

It follows that

a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n +  n(n-1)

so the left side is

2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n

Also recall that

\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}

so that the right side is

b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)

Solve for c.

2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}

Now, the numerator increases more slowly than the denominator, since

\dfrac{d}{dn}(2n(n-1)) = 4n - 2

\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2

and for n\ge5,

2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2

This means we only need to check if the claim is true for any n\in\{1,2,3,4\}.

n=1 doesn't work, since that makes c=0.

If n=2, then

c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0

If n=3, then

c = \dfrac{12}{2^3 - 6 - 1} = 12

If n=4, then

c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N

There is only one value for which the claim is true, c=12.

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2 years ago
Simplify square root of 324
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