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Stels [109]
3 years ago
11

Help meh sister... HELP MY POINTS THOOO :

Mathematics
2 answers:
Ugo [173]3 years ago
5 0

Answer:

2/3 times 42 is 28

Step-by-step explanation:

divide 2 and three = 0.6667

then times that by 42 and get 28

kolbaska11 [484]3 years ago
4 0

Answer:

Step-by-step explanation:

1.28

2.10 5/6

3.14 2/3

4. 20 1/4

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The data below shows the average number of text messages a group of students send per day. what is a histogram that represents t
Rainbow [258]
To draw a histogram, you need two axes. The horizontal axis is the day and the vertical axis is the number of text messages. For the first day, there are 20 text messages. This is represented by a bar starting from the horizontal axis labeled Day 1 going up to the point where the vertical axis indicates 20 text messages. This is done for the rest of the data. Afterwards, a line is drawn from the vertex of the axes connected to the top of the first bar. Then from the top of the first bar, a line connects it to the top of the next bar. This is done for all the bars before producing a histogram.
6 0
3 years ago
What is the selling price if the original cost is $145 and the markup is 150%? PLEASE HELP!! :(
Sergeu [11.5K]
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3 0
3 years ago
Help me i need help with my homework also i took a picture of my homerwork pls help
ICE Princess25 [194]

Answer:

123

Step-by-step explanation:

6 0
4 years ago
Read 2 more answers
Determine the measures of the angles of MNO if the measures of the angles are in the ratio 2:4:6
saul85 [17]
2:4:6 =2x :4x :6x
2x -1st angle
4x -2d one
6x - 3d one
2x+4x+6x=180 because MNO is a triangle
12x=180
x=180/12=15

2*15=30° -1st angle
4*15= 60° -2d one
6*15=90° - 3d one


4 0
3 years ago
Consider a particle that moves through the force field F(x, y) = (y − x)i + xyj from the point (0, 0) to the point (0, 1) along
sleet_krkn [62]

The work done by \vec F is

\displaystyle\int_C\vec F\cdot\mathrm d\vec r

where C is the given curve and \vec r(t) is the given parameterization of C. We have

\mathrm d\vec r=\dfrac{\mathrm d\vec r}{\mathrm dt}\mathrm dt=k(1-2t)\,\vec\imath+\vec\jmath

Then the work done by \vec F is

\displaystyle\int_0^1((t-kt(1-t))\,\vec\imath+kt^2(1-t)\,\vec\jmath)\cdot(k(1-2t)\,\vec\imath+\vec\jmath)\,\mathrm dt

=\displaystyle\int_0^1((k-k^2)t-(k-3k^2)t^2-(k+2k^2)t^3)\,\mathrm dt=-\frac k{12}

In order for the work to be 1, we need to have \boxed{k=-12}.

3 0
3 years ago
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