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Gnom [1K]
3 years ago
12

In the rectangular prism shown below, which lines are parallel?

Mathematics
2 answers:
olga_2 [115]3 years ago
8 0

Answer:

JN and LP...............

ValentinkaMS [17]3 years ago
7 0

Answer:

JM and KO

<h2 /><h2><u>HOPE IT HELPS</u></h2>

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Can you find the slope-intercept equation between each pair of coordinate points and type the correct code? Please remember to t
Annette [7]

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Step-by-step explanation:

4 0
3 years ago
Anyone think they could help me with this ?
topjm [15]

Answer:

10

Step-by-step explanation:

As we can see in hint through it we can identify that it is right angled triangle so here

AB = 6

FE = BC ( being corresponding sides of congruent triangle)

BC = 8

Now In triangle ABC

perpendicular (p) = 6

base (b) = 8

hypotenuse (h) = ?

By using Pythagoras theorem we get

h = √ p² + b²

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7 0
3 years ago
Let T:P3→P3 be the linear transformation such that T(â’2x2)=3x2+4x, T(0.5x+3)=â’3x2+3xâ’4, and T(2x2â’1)=â’3x+4. Find T(1), T(
balu736 [363]

It looks like we're told that

T(-2x^2)=3x^2+4x

T\left(\dfrac12x+3\right)=-3x^2+3x-4

T(2x^2-1)=-3x+4

We use the fact that T is linear to find T(1),T(x),T(x^2). First, we notice that

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

We also have

T\left(\dfrac12x+3\right)=T\left(\dfrac12x\right)+T(3)=\dfrac12T(x)+3T(1)

T(2x^2-1)=T(2x^2)-T(1)=2T(x^2)-T(1)

So once we find T(1), we can determine T(x) and T(x^2). We have

T(1)=-\left(T(-2x^2)+T(2x^2-1)\right)\implies T(1)=-3x^2-x-4

and using this we find

T(x)=12x^2+12x+16

T(x^2)=-\dfrac32x^2-2x

Then

T(ax^2+bx+c)=T(ax^2)+T(bx)+T(c)=aT(x^2)+bT(x)+cT(1)

T(ax^2+bx+c)=-\dfrac32\left(a-8b+2c\right)x^2-(2a-12b+c)x+16b-4c

8 0
4 years ago
How do you solve questions 6,8 and 10 in the picture?
tester [92]
The answer too #10 is 3
4 0
3 years ago
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