What will the length of segment I' J' be?

Mathematics

1 answer:

Ymorist [56]2 years ago

5 0

Answer:

5.333

Step-by-step explanation:

To find the length of dilated I' J' we need to dilate the shape.

To do this you multiply each coordinate by 8/3.

Ex: (x,y) ---> (8/3x, 8/3y)

H: (0,3) ---> (8/3*0, 8/3*3)

I: (1,1) ---> (8/3*1, 8/3*1)

J: (3,1) ---> (8/3*3, 8/3*1)

K: (2,5) ---> (8/3*2, 8/3*5)

Now we plot this. (Picture attached)

The new I' coordinate: (2.667, 2.667)

The new J' coordinate: (8, 2.667)

The new length would be 8 - 2.667 so we can find the distance between the two points.

Note we are subtracting the y coordinates and not x because x coordinates represent height and y represents length

8 - 2.667 equals 5.333

So the length of segment I' J' will be 5.333

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In an arithmetic sequence, t4 + t5 + t6 = 300, and t15 +t16+ t17 = 201. Find t18.

stiks02 [169]

In an arithmetic sequence:
Tn=t₁+(n-1)d

t₄=t₁+(4-1)d=t₁+3d
t₅=t₁+(5-1)d=t₁+4d
t₆=t₁+(6-1)d=t₁+5d

t₄+t₅+t₆=(t₁+3d) +(t₁+4d)+(t₁+5d)=3t₁+12d

Therefore:
3t₁+12d=300 (1)

t₁₅=t₁+(15-1)d=t₁+14d
t₁₆=t₁+(16-1)d=t₁+15d
t₁₇=t₁+(17-1)d=t₁+16d

t₁₅+t₁₆+t₁₇=(t₁+14d)+(t₁+15d)+(t₁+16d)=3t₁+45d

Therefore:
3t₁+45d=201 (2)

With the equations (1) and (2) we make an system of equations:

3t₁+12d=300
3t₁+45d=201

we can solve this system of equations by reduction method.

3t₁+12d=300
-(3t₁+45d=201)
-----------------------------
-33d=99 ⇒d=99/-33=-3

3t₁+12d=300
3t₁+12(-3)=300
3t₁-36=300
3t₁=300+36
3t₁=336
t₁=336/3
t₁=112

Threfore:

Tn=112+(n-1)(-3)
Tn=112-3n+3
Tn=115-3n

Now, we calculate T₁₈:

T₁₈=115-3(18)=115-54=61

Answer: T₁₈=61

5 0

3 years ago

Use the given endpoint R and the midpoint M of segment RS to find the coordinates of the other endpoint S. (Hint: Use the midpoi

zimovet [89]

Using the given endpoint R (8,0)and the midpoint M (4,-5) , the other endpoint S is (0,-10)

Explanation :

Use the given endpoint R and the midpoint M of segment RS

R (8, 0 ) and M (4, -5 )

Let 'S' be (x2,y2)

Apply the midpoint formula

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )\\$

Endpoint R is (x1,y1) that is (8,0)

Substitute the values and make it equal to M(4,-5)

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )\\\\(\frac{8+x_2}{2},\frac{0+y_2}{2} )=(4,-5)\\\\\frac{8+x_2}{2}=4\\8+x_2=8\\x_2=8-8=0\\x_2=0\\\\\\\frac{0+y_2}{2} =-5\\y_2=-10$