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kotegsom [21]
3 years ago
6

PLEASE HELP ME I BEG I REALLY NEED HELP WITH THIS PROBLEM!

Mathematics
1 answer:
lozanna [386]3 years ago
6 0
Area = length x width

x² + 12x = l * x


A.  

First, you factor x from x² + 12x

x^{2} +12x=x(x+12)

area = x (x+12)

area = (x+12) * x

Length = x + 12



B.

If x = 10,
l = x + 12 = 10 +12 = 22
length = 22
and
Area = x² + 12x = 10² + 12(10) = 100 + 120 = 220
area = 220
        

Hope this helps :)
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Z=5x- 11y <br> can someone please help
Feliz [49]

Answer:

z+11y/5 = x

Step-by-step explanation:

Since you are solving for x, you want to isolate the variable. You do that by adding 11y to both sides, which cancels out the -11y. From that you should have z+11y = 5x. To cancel out the 5x you need to divide both sides by 5 to get z+11y/5 = x. Hope this helps :)

3 0
3 years ago
NO LINKS!!!<br><br>QRSTU ~ FGCDE. Find FG, CD, and EF. ​
arsen [322]

Answer:

See below ~

Step-by-step explanation:

Given :

<u>QRSTU ~ FGCDE</u>

Finding the scale factor :

  • Take two corresponding sides in proportion
  • RS : GC
  • 40 : 12
  • <u>10 : 3</u>

Applying the scale factor to find the missing sides :

- FG :

  • QR : FG = 10 : 3
  • 30/FG = 10/3
  • FG = 30/10 x  3
  • FG = 3 x 3
  • <u>FG = 9</u>

- CD :

  • ST : CD = 10 : 3
  • 40/CD = 10/3
  • CD = 40/10 x 3
  • CD = 4 x 3
  • <u>CD = 12</u>

- EF :

  • UQ : EF = 10 : 3
  • 30/EF = 10/3
  • EF = 30/10 x 3
  • EF = 3 x 3
  • <u>EF = 9</u>
4 0
2 years ago
Read 2 more answers
Help me Please....................
MakcuM [25]

9514 1404 393

Answer:

  1.3363

Step-by-step explanation:

The basic idea here is to find an expression for the direction vector between a point on L1 and a point on L2. Then, solve for the points on L1 and L2 that make that vector perpendicular to both lines L1 and L2. (The dot product of direction vectors is zero.) The distance between the points found is the shortest distance between the lines.

__

Let P be a point on L1. Then the parametric equation for P is ...

  P = (6t, 0, -t) . . . . . . origin + t × direction vector

Let Q be a point on L2. The direction vector for L2 is given by the difference between the given points. It is (4-1, 1-(-1), 6-1) = (3, 2, 5). Then the parametric equation for Q is ...

  Q = (3s+1, 2s-1, 5s+1) . . . . (1, -1, 1) + s × direction vector

The direction vector for PQ is ...

  Q -P = (3s+1-6t, 2s-1, 5s+1+t)

The dot product of this and the two lines' direction vectors will be zero:

  (3s+1-6t, 2s-1, 5s+1+t)·(6, 0, -1) = 0 = 13s -37t +5 . . . perpendicular to L1

  (3s+1-6t, 2s-1, 5s+1+t)·(3, 2, 5) = 0 = 38s -13t +6 . . . perpendicular to L2

The solution to these equations is ...

  s = -157/1237

  t = 112/1237

Then (Q-P) becomes (94, -1551, 564)/1237, and its length is ...

  |PQ| = √(94² +1551² +564²)/1237 ≈ 1.3363

The distance between the two lines is about 1.3363 units.

8 0
3 years ago
A=1/2 (a+b)h work out A if a=2 b=7 h=3
BartSMP [9]

Step-by-step explanation:

A =1/2(a+b)h

  • 1/2(2+7)3
  • 1/2×27
  • 13.5

hope it helps.

6 0
2 years ago
Is the GCF of any two odd numbers always odd?
vampirchik [111]
Probably true

you can test it too
8 0
3 years ago
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