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Marina86 [1]
2 years ago
13

Someone help plzzzzz

Mathematics
1 answer:
Dovator [93]2 years ago
8 0

Answer:

for the first box the answer is 4x but I dont know the y intercept

Step-by-step explanation:

get 2 points the do y2-y1/x2-x1 then you get your x

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Two sides of a triangle
True [87]

step 1: open brackets n rmb to flip the signs

step 2 (optional): rearrange to make the equation clearer to add/minus

length of 3rd side:

(8x + 3) - (2x + 1) -(5x + 2)

= 8x + 3 - 2x -1 - 5x -2

= 8x -2x -5x +3 -1 -2

= x

Therefore, length of 3rd side is X cm.

---

If perimeter = 75cm,

8x +3 = 75

8x = 75-3

8x = 72

x = 9 cm

Therefore,

1st side = 2(9) +1

= 19 cm

2nd side = 5(9) +2

= 47 cm

3rd side = 9 cm

4 0
3 years ago
Marco wants to buy a new smartphone that costs $495, with some additional accessories. He has already saved $75. Write and solve
Dennis_Churaev [7]

Answer:

495-75=m m=420

Step-by-step explanation:

He needs to save 495 dollars, and hes already saved 75

and needs to find out how much more he needs to save to have 495 dollars

To do this, take 495 and subtract 75. This is how we get the inequality

495-75=m

From here it is simple, just subtract 75 from 495 to get 420

3 0
2 years ago
2x – 3y = 18<br> Solve for y.
Nataly [62]

Answer:

y= -6+2/3x

Step-by-step explanation:

2x-3y=18

Minus the 2x from both sides

-3y=18-2x

Then divide both sides by -1

-3y÷-1=18÷-1 -2x÷-1

Lastly divide both sides by 3

3y÷3= -18÷3+2x÷3

7 0
3 years ago
Write 4 numbers that round to 700,000 when rounded to the nearest hundred thousand
enot [183]
Well off the top of my head, (700,001),(700,002),(700,003), and (700,004) are 4 numbers that round to 700,00 when rounded to the nearest hundred thousand. But if you're looking for something a little more unique, then any number from 650,000 to 749,999 would round to to 700,000 when rounded to the nearest hundred thousand. I hope this helps!
4 0
3 years ago
Please answer the followings: W^mW^n=
Lunna [17]

Step-by-step explanation:

W^m=\underbrace{W\cdot W\cdot W\cdot...\cdot W}_{m}\\\\W^n=\underbrace{W\cdot W\cdot W\cdot...\cdot W}_{n}\\\\W^mW^n=\underbrace{(W\cdot W\cdot W\cdot...\cdot W)}_{m}\underbrace{(W\cdot W\cdot W\cdot...\cdot W)}_{n}\\\\=\underbrace{W\cdot W\cdot W\cdot...\cdot W}_{m+n}=W^{m+n}

5 0
3 years ago
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