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Solnce55 [7]
3 years ago
8

6 times some number equals 30

Mathematics
2 answers:
EastWind [94]3 years ago
5 0
6*1=6
6*2=12
6*3=18
6*4=24
6*5=30

So the answer is 5.

Hope this helps!
den301095 [7]3 years ago
4 0
Bro the answer to that is 5
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What geometric figure is the union of two non-collinear rays AB and AC?
Rufina [12.5K]
BC is the answer love... have fun on that I-Ready Diagnostic
6 0
3 years ago
Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL
Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

  • J(2;5); K(1;1)

             => JK = \sqrt{(1-2)^{2} + (1-5)^{2}  } = \sqrt{(-1)^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • J(2;5); L(5;2)

             => JL = \sqrt{(5-2)^{2} + (2-5)^{2}  } = \sqrt{3^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • K(1;1); L(5;2)

             =>  KL = \sqrt{(5-1)^{2} + (2-1)^{2}  } = \sqrt{4^{2}+1^{2}  } = \sqrt{1+16}=\sqrt{17}

In the triangle QNP, the sides can be calculate as following:

  • Q(-4;4); N(-3;0)

             => QN = \sqrt{[-3-(-4)]^{2} + (0-4)^{2}  } = \sqrt{1^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • Q (-4;4); P(-7;1)

   => QP = \sqrt{[-7-(-4)]^{2} + (1-4)^{2}  } = \sqrt{(-3)^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • N(-3;0); P(-7;1)

             =>  NP = \sqrt{[-7-(-3)]^{2} + (1-0)^{2}  } = \sqrt{(-4)^{2}+1^{2}  } = \sqrt{16+1}=\sqrt{17}

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0
3 years ago
Read 2 more answers
A circle has the center of (1,-5) and a radius of 5 determine the location of the point (4,-1)
Sliva [168]

"determine the location" or namely, is it inside the circle, outside the circle, or right ON the circle?

well, we know the center is at (1,-5) and it has a radius of 5, so the distance from the center to any point on the circle will just be 5, now if (4,-1) is less than that away, is inside, if more than that is outiside and if it's exactly 5 is right ON the circle.

well, we can check by simply getting the distance from the center to the point (4,-1).

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{center}{(\stackrel{x_1}{1}~,~\stackrel{y_1}{-5})}\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{-1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[4-1]^2+[-1-(-5)]^2}\implies d=\sqrt{(4-1)^2+(-1+5)^2} \\\\\\ d = \sqrt{3^2+4^2}\implies d =\sqrt{9+16}\implies d=\sqrt{25}\implies \stackrel{\textit{right on the circle}}{d = 5}

5 0
3 years ago
Points A(-2, 4), B(1, 3), C(4, -1) and D form a parallelogram. What are the coordinates of D?
Rudik [331]
Let point A(-2,4)  = A (X1,Y1)
      point B( 1,3  )= B (X2,Y2)
      point C(4,-1)  = C (X3,Y3)
and point D(?, ?) = D (X4,Y4)  We have to find this point

To find X4 we have to use the formula:
X2-X1=X3-X4

Now just plug in the numbers that correspond to the letters provided:
(1)-(-2)=(4)-(X4) ----> we don't know what X4 is yet, so we have to solve for it!
1+2=4-X4
3=4-X4
3-4=-X4
-1=-X4  divide both sides by -1
X4=1

Now we have to find Y4 using this formula:
Y2-Y1=Y3-Y4

Therefore,
(3)-(4)=(-1)-(Y4)
-1=-1-Y4
-1+1=-Y4
0=-Y4
So,
Y4=0


Now we have found the coordinates of the point D, which is (1,0)

Hope this helped!
8 0
3 years ago
Read 2 more answers
Please help 20 points!!
frutty [35]
Where is the question?
7 0
3 years ago
Read 2 more answers
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