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Nostrana [21]
3 years ago
7

BRAINLIEST!!! PLEASE HELP..!!

Mathematics
1 answer:
amm18123 years ago
7 0

Answer:

2nd choice:

x int: 11/6

y int: 11/2

Step-by-step explanation:

put 6x + 2y = 11 into y = mx+b form to find y intercept. (b stands for the y int)

you get y = -3x + 11/2.

now you can graph with this.

plug 0 in for y and solve for x to find x int

0 = -3x + 11/2 then solve

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I need help please?!!!!!
Ket [755]

Answer:

b)11

since 11^2 is 11*11 = 12

and when you take the square root of 121

we get 11

(in b- √11^2) , it cancels out the square

square root cancels square since they perform opposite functions

7 0
3 years ago
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can someone explain why it keeps saying my comments violate guidelines when i don’t say anything that shouldn’t be said?
Likurg_2 [28]

Answer:

Step-by-step explanation:

i think it’s just a glitch i keep getting those messages because i answered a question that got deleted

4 0
3 years ago
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A vegetable garden with an area of 200 square feet is to be fertilized. If the length of the garden is 1 foot less than three ti
charle [14.2K]
Ok so I like to go in steps with these questions- first draw a picture and identify your variables.

W=width
L= 3w-1

Now we know that length times width gets us area so we plug in our variables into the area equation.

200 = w(3w-1)

When you foil that equation you end up with a quadratic : 3w^2-w-200 = 0

Either factor that or use the quadratic formula to get
w= 8.33 and w= -8

Since you can't have a negative dimension you need to use 8.33 and plug it back into your length equation.

Final answer:

w= 8.33ft
l= 23.99ft

*Now I simplified the decimals a little bit so you end up with 199.8ft^2 for the area so just add a few decimals on here and there*
5 0
3 years ago
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Please help me with this problem!!! No Links!
DerKrebs [107]
The answer is 17! Hope it helps :)
4 0
3 years ago
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Prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)
EleoNora [17]
Start by multiplying each side of the inequality by \sqrt{x + 1} and simplifying:

2 \sqrt x + \frac{1}{\sqrt{x+1}}  \leq  2 \sqrt{x + 1}
(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})
2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)
2 \sqrt{x^2 + x} + 1 \leq 2x + 2
2 \sqrt{x^2 + x} \leq 2x + 1
\sqrt{x^2 + x} \leq x + \frac{1}{2}

From here, we can square both sides to get

x^2 + x \leq (x + \frac{1}{2})^2
x^2 + x \leq x^2 + x + \frac{1}{4}
0  \leq \frac{1}{4}, which is always true.
3 0
3 years ago
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