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katovenus [111]
3 years ago
11

59% of 127 i need the steps btw i am in middle school

Mathematics
2 answers:
Aleks [24]3 years ago
7 0

Answer:

74.93

Step-by-step explanation:

You just need to take 127 and multiply it by 59%:

127 · 0.59 = 74.93

Svetradugi [14.3K]3 years ago
6 0

Answer:

74.93

Step-by-step explanation:

59 x 127 /100

= 74.93

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2 years ago
Use all four operations and at least one exponent to write an expression that has a value of 100.
OleMash [197]

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Step-by-step explanation:

3 0
3 years ago
Write the linear inequality shown in the graph.
Korolek [52]
Fist, in the equation y=mx+b, b is the y-intercept. The y-intercept is the poin on the line that crosses the x-axis; the y-intercept is the value of yThese equations follow that format. 
Y=mx+b
y\geq 3x-4. <-----In this equation, the slope(m)=3 b= -4. 

On the graph, we can see that the line crosses the x-axis at y=-4. Knowing that, we can eliminate the answer choices with +4 in the inequality.


The next step, is to pick an (x,y) coordinate that is in the shaded region and plug it into the remaining 2 inequalities. Which ever inequality is true after you solve it, that is the correct answer.  
For example, I'll choose to plug in (-4,4) into the y\geq 3x-4. 
y\geq 3x-4. 
(4)\geq (3(-4))-4. 
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So, this statement is true becasue -16 is less than positive 4.  Therefore, the correct answer would be y\geq 3x-4. Hope that helped! Comment back with any further questions!


6 0
3 years ago
Martin earned the following scores on his last five tests.
Arada [10]
IQR= 18 is the interquartile range of his scores.
8 0
4 years ago
Which set of coefficients of the terms in the expansion of the binomial (x+y)^3 is correct ?
BigorU [14]
We are technically FOILing this out... with a power of 3.

(x+y)(x+y)(x+y)

So we can first factor out the first two "x+y"s.

( x^{2} +2xy + y^{2} ), multiplied by the last "x+y".

x^{3} +3 x^{2} y +3x y^{2} + y^{3}

Coefficients are the number that comes in front of a variable. 

In this case, 1 comes in front of x^{3}, 3 comes in front of x^{2} y, 3 comes in front of x y^{2}, and 1 comes in front of y^{3}.

Thus: 1, 3, 3, 1.
         Answer Choice A
4 0
3 years ago
Read 2 more answers
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