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ICE Princess25 [194]
3 years ago
12

Solve the problem for Y 3x + y = 7

Mathematics
1 answer:
sergejj [24]3 years ago
6 0

Answer:

not possible with info given

Step-by-step explanation:

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A sequence of translations maps ∆ABC to ∆A’B’C’.
Len [333]

Answer:

(a) Co-ordinate rule is x'=x+6 and y'=y-8

(b) Co-ordinates of B' and C' are (3,-8) and (5,-5) respectively.

Step-by-step explanation:

(a)

Here, the co-ordinates of A(-3,4) are translated to A'(3,-4).

For the co-ordinates A and A',  x'-x=3-(-3)=3+3=6 and y'-y=-4-4=-8

So, x value of A has shifted to right by 6 units and y value of A has shifted 8 units down.

Hence, the co-ordinate rule that maps ΔABC onto ΔA'B'C' is:

x'=x+6 and y'=y-8.

(b)

Using the co-ordinate rule, we can find the co-ordinates of B' and C'.

For B, x=-3 and y=0.

So, x' of B' is x'=x+6=-3+6=3

And, y' of B' is y'=y-8=0-8=-8.

Therefore, co-ordinates of B' are (3,-8).

For C, x=-1 and y=3.

So, x' of C' is x'=x+6=-1+6=5

And, y' of C' is y'=y-8=3-8=-5.

Therefore, co-ordinates of C' are (5,-5).

3 0
3 years ago
ine segment XY has endpoints X(–10, –1) and Y(5, 15). To find the y-coordinate of the point that divides the directed line segme
stiks02 [169]
X=(-10,-1)=(Xx,Yx)→Xx=-10, Yx=-1
Y=(5,15)=(Xy,Yy)→Xy=5,Yy=15
y=?
ratio=5:3→r=5/3
y=(Yx+rYy) / (1+r)
y=[-1+(5/3)15] / (1+5/3)
y=[-1+(5*15)/3] / [(3+5)/3]
y=(-1+75/3) / (8/3)
y=(-1+25) / (8/3)
y=(24) / (8/3)
y=24*(3/8)
y=72/8
y=9

Asnwer: T<span>he y-coordinate of the point that divides the directed line segment XY in a 5:3 ratio is y=9</span>
3 0
3 years ago
Read 2 more answers
Using the method of mathematical induction to prove that equalities are true for values ​​of n indicated:
Dima020 [189]
2^2+4^2+6^2+...(2n)^2=\frac{2n(n+1)(2n+1)}{3};\ n\geq1\\\\chek\ for\ n=1:\\L=2^2=4;\ R=\frac{2\cdot1(1+1)(2\cdot1+1)}{3}=\frac{2\cdot2\cdot3}{3}=4\\L=R\\-----------------------\\&#10;assumption\ for\ n=k\\2^2+4^2+6^2+...+(2k)^2=\frac{2k(k+1)(2k+1)}{3}\\-----------------------\\thesis\ for\ n=k+1\\2^2+4^2+6^2+...+(2k)^2+[2(k+1)]^2=\frac{2(k+1)(k+1+1)[2(k+1)+1]}{3}\\-----------------------
proff:\\L=2^2+4^2+6^2+...+(2k)^2+(2k+2)^2=\frac{2k(k+1)(2k+1)}{3}+(2k+2)^2\\\\=\frac{(2k^2+2k)(2k+1)}{3}+\frac{3(2k+2)^2}{3}=\frac{4k^3+2k^2+4k^2+2k+3(4k^2+8k+4)}{3}\\\\=\frac{4k^3+6k^2+2k+12k^2+24k+12}{3}=\boxed{\frac{4k^3+18k^2+26k+12}{3}}\\\\R=\frac{2(k+1)(k+1+1)[2(k+1)+1]}{3}=\frac{(2k+2)(k+2)(2k+2+1)}{3}\\\\=\frac{(2k^2+4k+2k+4)(2k+3)}{3}=\frac{(2k^2+6k+4)(2k+3)}{3}=\frac{4k^3+6k^2+12k^2+18k+8k+12}{3}\\\\=\boxed{\frac{4k^3+18k^2+26k+12}{3}}\\\\L=R
4 0
3 years ago
What is the slope-intercept form of 9x + 3y = 15?
Sholpan [36]

Answer:

b) y= 3x - 5

Step-by-step explanation:

9x-15 ÷3 = 3x - 5

3y ÷ 3 = y

8 0
3 years ago
If the probability that a student is on-time to class is 0.1 and the probability he is on-time or marked tardy on the attendance
kompoz [17]

T Tardy– Student tardy for less than 30 minute with no valid excuse. W See Unexcused Absences (above). Z Excused Tardy– Student tardy to class but with an acceptable explanation, i.e. doctor/dentist.
3 0
3 years ago
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