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lions [1.4K]
3 years ago
8

Point R is at (2, 1.2) and Point T is at (2, 2.5) on a coordinate grid. The distance between the two points is ____. (Input numb

ers and decimal point only, such as 8.2.)
Mathematics
1 answer:
Nataly [62]3 years ago
3 0
You could graph this, but it is simpler in this case to do it mathematically. 

Since the x and values are the same (implying undefined slope) you can subtract the smaller y value from the larger y value. 2.5-1.2=1.3. 
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Lydia figures it should be easy to carry her golf clubs in her new golf bag to be sure she decides to check the total weight the
balu736 [363]
Since 12.3 is kind of a margin that you have to fit by, the total weight of the golf clubs must be lesser than or equal to 12.3. So, that's your expression:
0.8x \leq 12.3
x represents the number of golf clubs you have.

If this doesn't make sense, feel free tell me!
6 0
3 years ago
Please help me with expansion and factorization of quadratic expressions.<br> a) -4x^2 + 2x^2
Flura [38]

Answer:-2x^2(2-1)

Step-by-step explanation:

-4x^2 + 2x^2=-2x^2(2 - 1)

8 0
3 years ago
Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess
Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2

4 0
3 years ago
Read 2 more answers
Compare. Write &lt;, &gt;, or=.<br>11. V8 +3( ) 8+ V3​
Levart [38]

Answer:

=

Step-by-step explanation:

v8 + 3 = 8 + v3

-3 -3

________________

v8 = 5 + v3

-v3 -v3

___________

v5 = 5

8 0
3 years ago
What is the range of data in this box plot? what is the median?
jeka57 [31]
Hello!

The range is the lowest number subtracted from the highest number

The lowest number is 1
The highest number is 9

9 - 1 = 8

The range is 8
------------------------------------------------------------------------------------------------------
To find the median you look at the line that separates the box plot into two sections

The median is 4

Hope this helps!
8 0
4 years ago
Read 2 more answers
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