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Black_prince [1.1K]
2 years ago
12

3x – 57 + x = 63 solve for x

Mathematics
2 answers:
e-lub [12.9K]2 years ago
7 0

Answer:

x=30

Step-by-step explanation:

ANEK [815]2 years ago
5 0
The answer is 30
 x=30
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On a trip, you had to change your money from dollars to British pounds. You got 560560 pounds for 800800 dollars. How many pound
Vlad [161]
First you need to divide 800,800 by 300,300 because you need to know how these two amounts are related, in order to know how the amounts of pounds you are converting are related.
800,800÷300,300= 2.66666.... (recurring)
If $300,300 goes into $800,800 2.66666 times, then the amount of pounds you want to find goes into £560,560 2.66666 times as well. Let's write this as an equation to solve for x (where x is the amount of pounds you will obtain for $560,560):
2.66666x = 560,560
Divide both sides by 2.66666
x = 210,210
Therefore, you will get £210,210 for $300,000
7 0
3 years ago
Pls solve will mark brainiest, five stars and will give a thanks
Goryan [66]

Answer:

8 times

Step-by-step explanation:

3/4 n = 6

Multiply each side by 4/3 to isolate n

4/3 * 3/4 n = 6*4/3

n = 8

8 0
3 years ago
Read 2 more answers
A line passes through the points (–1, –5) and (4, 5). The point (a, 1) is also on the line.
IgorC [24]

Answer:

a = 2

Step-by-step explanation:

Let's find the equation of the line using the 2 points given.

Let's call -1 as x_1 and -5 as y_1

also, 4 as x_2 and 5 as y_2

The equation of line is :

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

<em>Let's plug the points and get the equation of the line:</em>

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\y+5=\frac{5+5}{4+1}(x+1)\\y+5=2(x+1)\\y+5=2x+2\\y=2x-3

Now, to find a, we substitute a in x and 1 in y of the equation of the line we just got:

y=2x-3\\1=2(a)-3\\1=2a-3\\2a=4\\a=\frac{4}{2}=2

4 0
3 years ago
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HELP ME ASAP LAST PROBLEM
igomit [66]
The rate of change is -2 :)

5 0
2 years ago
Read 2 more answers
The volume
Sedaia [141]
\bf \begin{array}{cccccclllll}&#10;\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\&#10;\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\&#10;y&=&{{ k}}&\cdot&x&#10;&&  y={{ k }}x&#10;\end{array}\\ \quad \\&#10;

and also

\bf \begin{array}{llllll}&#10;\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\&#10;\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\&#10;y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}&#10;&&y=\cfrac{{{  k}}}{x}&#10;\end{array}&#10;

now, we know that V varies directly to T and inversely to P simultaneously
thus\bf V=T\cdot \cfrac{k}{P}

so     \bf V=T\cdot \cfrac{k}{P}\qquad &#10;\begin{cases}&#10;V=42\\&#10;T=84\\&#10;P=8&#10;\end{cases}\implies 42=\cfrac{84k}{8}\implies 4=k&#10;\\\\\\&#10;V=\cfrac{4T}{P}\qquad now\quad &#10;\begin{cases}&#10;V=74\\&#10;P=10&#10;\end{cases}\implies 74=\cfrac{4T}{10}\implies 185=T
7 0
3 years ago
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