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3 years ago

5u-2u=..................................................

Mathematics

1 answer:

MAVERICK [17]3 years ago

8 0

It will be 3u good luck

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Consider the following data for 120 mathematics students at a college concerning the languages French, German, and Russian:

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Answer:

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3 years ago

There are 28 students in a class. the teacher separates them into 7 equal groups. which equation can be used to find s, the numb

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S = total # of students / #of students in each group
So you can do, s= t/g
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3 years ago

Please help me with the question

Anna35 [415]

Answer: 70

Step-by-step explanation: if you look at it hard enough they both look the same, this is called a vertical angle, I'm down for more questions lol. you keep on getting on my fyp

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3 years ago

find the equation of the perpendicular bisector of segment ab with endpoints a=2,2 and b=-4,6 if the perpendicular bisector inte

aleksklad [387]

Equation of perpendicular bisector is , $y=\frac{3}{2} x+\frac{11}{2}$

Length of both pa and pb are $\sqrt{\frac{65}{4} }$ unit.

Slope of segment ab is, $\frac{6-2}{-4-2} =-\frac{2}{3}$

So, slope of perpendicular bisector will be, negative of reciprocal of slope of segment ab.

Therefore, slope of perpendicular bisector is, $\frac{3}{2}$

Since, perpendicular bisector passes through the mid point of segment ab.

So, coordinate of mid point of segment ab is,

$(\frac{2-4}{2},\frac{2+6}{2} )=(-1,4)$

Thus, perpendicular bisector of ab passing through (- 1, 4) and have slope 3/2.

So, equation of angle bisector is,

$y=\frac{3}{2} x+\frac{11}{2}$

Since, the perpendicular bisector intersects the y-axis at point p

So, coordinate of point p is (0, 11/2)

By applying distance formula,

$pa=\sqrt{(2-0)^{2}+(2-11/2)^{2} }\\\\ =\sqrt{4+\frac{49}{4} }\\\\ =\sqrt{\frac{65}{4} }$

$pb=\sqrt{(-4-0)^{2}+(6-11/2)^{2} }\\\\ =\sqrt{16+\frac{1}{4} }\\\\ =\sqrt{\frac{65}{4} }$

Learn more,

https://brainly.in/question/31510165

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3 years ago

Write 5 ratios that are equivalent to 3 sodas sold for<br> every 10 bags of chips!? :)

Vitek1552 [10]

Umm this one is a little challenging I’ll be back

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3 years ago