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3 years ago

What is 10/18 equivalent to

Mathematics

1 answer:

a_sh-v [17]3 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°

4 0

3 years ago

The measures of ∠x and ∠y are equal. Which is the value of m?

morpeh [17]

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.

4 0

3 years ago

Read 2 more answers

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

6 0

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

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