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2 years ago

Help a girl outttt Pleaseeee

Mathematics

1 answer:

Tomtit [17]2 years ago

5 0

$⟹3x+5 = -x-7$

$⟹3x+x=-7-5$

$⟹4x=-12$

$⟹3-5x=-9+x$

$⟹3+9=x+5x$

$⟹12=6x$

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I need help o really don't understand

aleksklad [387]

1/3+2/3=1 is the answer

7 0

2 years ago

Read 2 more answers

Points A and B are on the different sides of a line 1, the distance between point A and the line lis 10 in, the distance between

givi [52]

Answer: Distance between O and the line 1 is 3 inches.

Explanation:

Since we have given that

Distance between point A and the line 1 = 10 inches

Distance between point B and the line 1 = 4 inches

Total length of segment AB is given by

$10+4=14\ inches$

Since O is midpoint of the line segment AB.

so, AO is given by

$\frac{14}{2}=7\ in.$

Now, Distance between O and the line 1 is given by

$10-7=3\ in.$

Hence, Distance between O and the line 1 is 3 inches.

7 0

3 years ago

Which represents the solution to the absolute value equation? 3|2x+4|-1=11

larisa86 [58]

The first step for absolute value equations is to isolate the expression contained within the absolute value bars:

3|2x+4|-1 = 11
3|2x+4| = 12
|2x+4| = 4

so |2x+4| is 4 units away from 0 on a number line, but we don't know in which direction -- negative or positive? you'll have two answers.

2x+4 = 4

AND

2x+4 = -4

solve both of those two step equations and you'll get

x = 0

AND

x = -4

so 0 and -4 are your solutions.

5 0

2 years ago

Max is 6 years older than his brother. If Max is M years old, then his brother is ____ years old.

Doss [256]

Max' brother is M-6 years old.

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

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