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3 years ago

Solve this equation a{6a-2a}=b{2a-3b}

1 answer:

4 0

The "solution" of the equation would be a pair of numbers ... one
for 'a' and one for 'b' ... that make the equation a true statement.

If the equation has two variables in it, as this one does, then
you need two equations to find the solution.

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Lake Michigan is the fifth largest lake

nikklg [1K]

Yes with a 58,016 Square Kilometers

6 0

3 years ago

Read 2 more answers

Line segment JK measures 17cm. Segments JK was translated 4 units right and 8 units up for for J'K'. What is the measurement of

castortr0y [4]

Answer:

Step-by-step explanation:

8 0

2 years ago

A truck driver can travel 560 miles on 28 gallons of gas. How far can he travel on 35 gallons of gas?

Shtirlitz [24]

560 / 28 = 20....so the driver travels 20 miles per gallon

on 35 gallons......35 * 20 = 700 miles <==

or u can also do it this way...

560/28 = x / 35....560 miles to 28 gal = x miles to 35 gal
cross multiply because this is a proportion
(28)(x) = (560)(35)
28x = 19600
x = 19600/28
x = 700 miles <==

so u can either do it by finding the unit rate or by making a proportion. either way will result in the same answer.

8 0

3 years ago

HELP!!!!!!! images below

Umnica [9.8K]

Step-by-step explanation:

me manda as respostas em português

5 0

3 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

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