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pychu [463]
3 years ago
9

Mai’s backpack weighs 9 pounds. Han’s backpack weighs 8.3 pounds. Priya’s backpack weighs 7.75 pounds, and Kiran’s backpack weig

hs 6.125 pounds. Select all the true statements.
Group of answer choices

Han’s backpack is 2.175 pounds heavier than Kiran’s.

Mai’s backpack is 2.25 pounds heavier than Priya’s.

Han’s backpack is 16.05 pounds heavier than Priya’s.

Kiran’s backpack and Han’s backpack together weigh 14.3125 pounds.

The backpacks weigh 31.175 pounds together.
Mathematics
2 answers:
ale4655 [162]3 years ago
8 0

Answer: Han’s backpack is 16.05 pounds heavier than Priya’s.

Step-by-step expl

Alex3 years ago
7 0
Han’s backpack is 2.175 pounds heavier than Kiran’s.
The rest of the choices don’t add up.
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Step-by-step explanation:

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How do I convert 1 5/6 into a decimal??????
zavuch27 [327]

Answer:

1.8(3) (three is a repeating number)

Step-by-step explanation:

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3 years ago
Read 2 more answers
What is true about the completely simplified sum of the polynomials 3x^2y^2-2xy^5 and -3x^2y^2 + 3x^2y^2 + 3x^4y?
WINSTONCH [101]

Given:

The polynomials are:

3x^2y^2-2xy^5

-3x^2y^2+3x^2y^2+3x^4y

To find:

The completely simplified sum of the polynomials.

Solution:

We have,

3x^2y^2-2xy^5

-3x^2y^2+3x^2y^2+3x^4y

The sum of given polynomials is:

Sum=3x^2y^2-2xy^5-3x^2y^2+3x^2y^2+3x^4y

Sum=-2xy^5+3x^4y+3x^2y^2

Therefore, the sum of the given polynomials is -2xy^5+3x^4y+3x^2y^2. It is a polynomial with degree 6 and leading coefficient -2.

5 0
3 years ago
Greg is trying to solve a puzzle where he has to figure out two numbers, x and y. Three less than two-third of x is greater than
Fittoniya [83]
Inequation 1: 

\frac{2}{3}x-3 \geq y

to plot the pairs (x, y) for which the inequation holds, draw the line y=\frac{2}{3}x-3

then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.

we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.

inequation 2: 

y+ \frac{2}{3}x\ \textless \ 4

y\ \textless \ - \frac{2}{3} x+ 4



draw the lines 

i)  y=\frac{2}{3}x-3          use points (0, -3),  (3, -1)

ii)y=- \frac{2}{3} x+ 4       use points ( 0, 4),   (3, 2)


let's use the point P(3, 3) to see what region of the lines need to be coloured:

\frac{2}{3}x-3 \geq y  ; 
\frac{2}{3}(3)-3 \geq 3
2-3 \geq 3, not true so we color the region not containing this point


y+ \frac{2}{3}x\ \textless \ 4
(3)+ \frac{2}{3}(3)\ \textless \ 4
3+ 5\ \textless \ 4 not true, so we color the region not containing the point (3, 3)

The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.



4 0
3 years ago
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