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Taya2010 [7]
2 years ago
5

I posted this again with more points I'll mark the first right answer with brainliest please help no explanation required

Mathematics
1 answer:
Rainbow [258]2 years ago
8 0

?Step-by-step explanation:

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Tell whether (1,6) is a solution of y = x + 7
andrezito [222]

Answer:

no it is not a solution.READ THE WHOLE EXPLANATION! IT WILL HELP WITH THE OTHER PROBLEMS!!!!

Step-by-step explanation:

when something like (1,6) is written consider it as (x,y). in this case, 1 is considered as x, and 6 is considered as y. if you substitute these number for x and x and y, (x =1 and y =6), you'll see that the equation will go from y=x +7, to 6=1+7. Which is not true.

5 0
2 years ago
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Solve the equation x/2=x/3+1/2​
Mice21 [21]

Answer: x=3

Step-by-step explanation:

Move all x terms to the same side- x/2-x/3-1/2

Multiply to get common denominator- 3x/6-2x/6=1/2

Subtract 3x/6 and 2x/6 - x/6=1/2

Simplify - x=3

7 0
3 years ago
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If you need 192 beads to make a necklace and bracelet and it takes 5 times more beads to make necklace than bracelet. How many b
Rasek [7]
192 = total beads

necklace + bracelet = 192

Let x equal the number of beads 

bracelet = 5x
necklace = x

5x + x = 192 

6x = 192

x = 32

It takes 32 beads to make a necklace

3 0
2 years ago
-<br> Which expression is equivalent to 3x+10-x+12
stiv31 [10]

Answer:

Step-by-step explanation:

3x-x+= 2x

10+ 12= 22

Answer - 2x+22

5 0
2 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%20%7B%7D%5E%7B2%7D%20-%204%20%7D%7B%20%5Csqrt%7Bx%20%7B%7D%5E%7B2%7D%20-%206x%2
Nimfa-mama [501]

First of all, we can observe that

x^2-6x+9 = (x-3)^2

So the expression becomes

\dfrac{x^2-4}{\sqrt{(x-3)^2}} = \dfrac{x^2-4}{|x-3|}

This means that the expression is defined for every x\neq 3

Now, since the denominator is always positive (when it exists), the fraction can only be positive if the denominator is also positive: we must ask

x^2-4 \geq 0 \iff x\leq -2 \lor x\geq 2

Since we can't accept 3 as an answer, the actual solution set is

(-\infty,-2] \cup [2,3) \cup (3,\infty)

7 0
2 years ago
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