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vagabundo [1.1K]
3 years ago
6

Write an equation that represents the line. Use exact numbers

Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
7 0
Y=2/3x+2/3 bc slope is 2/3 and 4-2/4-1=2/3
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Someone please help me
Kitty [74]

Answer:

0.45, 4, 2, 1.5

Step-by-step explanation:

14/3 = 4.66667

Numbers less than 14/3 are 0.45, 4, 2, 1.5.

8 0
3 years ago
What is the approximate distance between the points (-3, -4) and (-8, 1) on a coordinate grid
tatyana61 [14]
\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ -3 &,& -4~) 
%  (c,d)
&&(~ -8 &,& 1~)
\end{array}\\\\\\ %  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{[-8-(-3)]^2+[1-(-4)]^2}\implies d=\sqrt{(-8+3)^2+(1+4)^2}
\\\\\\
d=\sqrt{25+25}\implies d=\sqrt{2\cdot 25}\implies d=\sqrt{2\cdot 5^2}\implies d=5\sqrt{2}
4 0
3 years ago
Read 2 more answers
Directions: Complete all 3 questions. Make sure you include a graph, work and conclusion.
Simora [160]

Prove that the quadrilateral whose vertices are I(-2,3), J(2,6), K(7,6), and L(3, 3) is a rhombus.

I think in these problems the first step is to express each side as a vector.  A vector is the difference between points.  When two sides have the same vector (or negatives) it means they're parallel and congruent.  So in a rhombus IJKL the vectors IJ and LK should be the same, as should JK and IL.  That much assures a parallelogram; we check IJ and JK are congruent to complete the crowing of the rhombus.

Let's calculate these vectors:

IJ = J - I = (2,6) - (-2,3) = (2 - -2, 6 - 3) = (4, 3)

LK = K - L = (7, 6) - (3, 3) = (4, 3)

IJ = LK, so far so good

(Note: If you haven't got to vectors yet you can just show the two sides are the same length, 5, and have the same slope, 3/4, both of which can be read off the vectors.)

JK = K - J = (7,6) - (2,6) = (5,0)

IL = L - I = (3, 3) - (-2, 3)  = (5, 0)

Those are the same too.    

Now we have to show IJ ≅ JK

The length of IJ is the cliche √4²+3² = 5, the same as JK, so IJ ≅ JK

We showed all four sides are congruent and we have two pair of parallel sides, so we have a rhombus.

8 0
4 years ago
Find the dimensions of a closed right circular cylindrical can of smallest surface area whose volume is 128pi cm^3
Ratling [72]
The surface area of a cylindrical can is equal to the sum of the area of two circles and the body of the cylinder: 2πr2 + 2πrh. volume is equal to π<span>r2h.

V = </span>π<span>r2h = 128 pi
r2h = 128 
h = 128/r2

A = </span><span>2πr2 + 2πrh
</span>A = 2πr2 + 2πr*(<span>128/r2)
</span>A = 2πr2 + 256 <span>π / r
</span><span>
the optimum dimensions is determined by taking the first derivative and equating to zero.

dA = 4 </span>πr - 256 <span>π /r2 = 0
r = 4 cm
h = 8 cm
</span><span>

</span>
4 0
3 years ago
Y= -0.25x + 4.7<br> Y=4.9x - 1.64
Andreas93 [3]

Answer:

The answer would be (1.23, 4.39)

Step-by-step explanation:

Because they are both equal to y, we can set the equations equal to each other and then solve.

4.9x - 1.64 = -0.25x + 4.7

5.15x - 1.64 = 4.7

5.15x = 6.34

x = 1.23

Now that we have the value for x, we can plug into either equation and find y.

y = -0.25x + 4.7

y = -0.25(1.23) + 4.7

y = -.31 + 4.7

y = 4.39

3 0
4 years ago
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