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Arada [10]
2 years ago
6

Name three synonyms for "Solution": _____________,___________,______________

Mathematics
1 answer:
Damm [24]2 years ago
6 0

Answer:

1 quick fix            2 solving               3 result

Step-by-step explanation:

hope it helps

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taurus [48]
8. C

9. C

10. A

Hope this helped!
5 0
3 years ago
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What is the slope of a line with coordinates (5,3) and (1,2)?<br><br> A (1,4)<br> B 4<br> C (1,2)
Brilliant_brown [7]

Answer:

Slope (m) =

ΔY

ΔX

=

1

4

= 0.25

θ =  

arctan( ΔY ) + 180°

ΔX

= 194.03624346793°

ΔX = 1 – 5 = -4

ΔY = 2 – 3 = -1

Distance (d) = √ΔX2 + ΔY2 = √17 = 4.1231056256177

Equation of the line:

y = 0.25x + 1.75

or

y =  

1 x

4

+  7

4

When x=0, y = 1.75

When y=0, x = -7

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Which ratios are less than 8 to 10? Check all that apply.<br> 6/20<br> 3:5<br> 50/100<br> 13:15
poizon [28]

9514 1404 393

Answer:

  6/20, 3:5, 50/100 are all less than 8/10

Step-by-step explanation:

It can help to give all the ratios a common denominator. (We'll use 100, and fudge the last one a bit.)

  8/10 = 80/100 . . . . our reference value

__

  6/20 = 30/100 . . . less than 8/10

  3/5 = 60/100 . . . . less than 8/10

  50/100 . . . . . . . . . less than 8/10

  13/15 ≈ 86.7/100 . . . more than 8/10

6 0
3 years ago
What type of display would you use to show the number of wins the school volleyball team had from 2000 to 2005
Agata [3.3K]
I'd use line chart, percentage or real number. Hope this helps
6 0
3 years ago
Find the area of the region in the first quadrant bounded on the left by the ​y-​axis, below by the line y equals one third x co
bonufazy [111]

Answer:

The bounded area  is: \frac{73}{6}\approx 12.17

Step-by-step explanation:

Let's start by plotting the functions that enclose the area, so we can find how to practically use integration. Please see attached image where the area in question has been highlighted in light green. The important points that define where the integrations should be performed are also identified with dots in darker green color. These two important points are: (2, 6) and (3, 1)

So we need to perform two separate integrals and add the appropriate areas at the end. The first integral is that of the difference of function y=x+4 minus function y=(1/3)x , and this integral should go from x = 0 to x = 2 (see the bottom left image with the area in red:

\int\limits^2_0 {x+4-\frac{x}{3} } \, dx =\int\limits^2_0 {\frac{2x}{3} +4} \, dx=\frac{4}{3} +8= \frac{28}{3}

The next integral is that of the difference between y=-x^2+10 and the bottom line defined by: y = (1/3) x. This integration is in between x = 2 and x = 3 (see bottom right image with the area in red:

\int\limits^3_2 {-x^2+10-\frac{x}{3} } \, dx =-9+30-\frac{3}{2} -(-\frac{8}{3} +20-\frac{2}{3} )=\frac{39}{2} -\frac{50}{3} =\frac{17}{6}

Now we need to add the two areas found in order to get the total area:

\frac{28}{3} +\frac{17}{6} =\frac{73}{6}\approx 12.17

8 0
3 years ago
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