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Soloha48 [4]
3 years ago
12

PLEASE HELP!!

Mathematics
2 answers:
Andre45 [30]3 years ago
8 0

Answer:

I think the answer is c

Step-by-step explanation:

Because it says at most? am I right

Afina-wow [57]3 years ago
5 0

Answer: 2w+2 x (4w) _< 106

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

  • <em>Perimeter of a rectangle: 2w +2L </em>

Where:

w: width

L : length

So, if the length of a rectangle is four times its width, it means that  

L = 4 w

Replacing with the new value of "L":

Perimeter: 2w + 2(4w)

The perimeter is at most 106, at most means equal or lower than. So:

2w+2 x (4w) _< 106

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Please help, will give brainliest!<br>(please explain how as well)​
vovikov84 [41]

Answer:

C

Step-by-step explanation:

Since GJ bisects ∠ FGH , then ∠ FGJ = ∠ JGH = x + 14

∠ FGH = ∠ FGJ + ∠ JGH , substitute values

4x + 16 = x + 14 + x + 14 = 2x + 28 ( subtract 2x from both sides )

2x + 16 = 28 ( subtract 16 from both sides )

2x = 12 ( divide both sides by 2 )

x = 6

Thus

∠ FGJ = x + 14 = 6 + 14 = 20° → C

4 0
2 years ago
Let f be the function given by f(x=sin(5x pi/4
sasho [114]
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3 years ago
Convert 900% to decimal​
Inessa [10]

Answer:

9

Step-by-step explanation:

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4 0
3 years ago
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In a candy bar, 165 calories are from sugar. There are 250 calories in the candy bar. What percent of the calories are from suga
stiv31 [10]
165/200=33/40
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Hope this helps
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3 years ago
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morpeh [17]
The first step for solving this expression is to reduce the fraction with 11.
\sqrt{ \frac{5 x^{7} y^{6}  }{ x^{11}  y^{8} } }
Now reduce the fraction with x^{7}.
\sqrt{ \frac{5 y^{6} }{ x^{4} y^{8}  } }
Reduce the fraction one more time with y^{6}.
\sqrt{ \frac{5}{ x^{4} y^{2}  } }
To take the root of a fraction,, we must take the root of the numerator and denominator separately. This rule will change the expression to the following:
\frac{ \sqrt{5} }{ x^{2} y}
Since we cannot simplify this fraction any further,, the final answer is going to be \frac{ \sqrt{5} }{ x^{2} y}.
Let me know if you have any further questions.
:)
3 0
3 years ago
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