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cupoosta [38]
3 years ago
15

How to find the value of x in each figure?​

Mathematics
2 answers:
mezya [45]3 years ago
7 0

You need to multiply the length and the width of the shape.

Maksim231197 [3]3 years ago
5 0

There are many different formulas, it depends on what shape it is. So what shape do you need to solve for?

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In the figure below, MN | KL and are intersected by transversals ML and KN M 1551 What is the measure of , XMN?
Schach [20]

Answer:

∠ XMN = 50°

Step-by-step explanation:

∠ XNM and ∠ XKL are alternate angles and are congruent, then

∠ XNM = 35°

The sum of the 3 angles in a triangle = 180° , so

∠ XMN = 180° - (95 + 35)° = 180° - 130° = 50°

5 0
3 years ago
I need help, also give explantion please
HACTEHA [7]

Answer:

\huge  \: 8 {x}^{3}  \times 3 {x}^{9}  = 24 {x}^{12}

3 0
3 years ago
Value of x in the given figure​
ololo11 [35]

Answer:

\sf\longmapsto \: x = 3

Step-by-step explanation:

\sf\longmapsto \: 6x  - 7 = 4x - 1

\sf\longmapsto \: 6x - 4x =  - 1 + 7

\sf\longmapsto \: 2x = 6

\sf\longmapsto \: x = 6 \div 2

\sf\longmapsto \: x = 3

6 0
2 years ago
Read 2 more answers
Using the technique in the model above, find the missing sides in this 30°-60°-90° triangle.
lina2011 [118]
For this case what you should do is use the following trigonometric relationship:
 sin (x) = C.O / h
 Where
 x: angle
 C.O: opposite leg
 h: hypotenuse
 Substituting the values we have:
 sen (60) = long / h
 sen (60) = 3 / h
 h = 3 / sin (60)
 h = 3.46
 Answer:
 h = 3.46
7 0
3 years ago
Read 2 more answers
Write an equation of the parabola that opens up whose vertex (−1, 2) is 3 units from the focus. (Vertex form or Parabola Form is
TEA [102]

Answer:

y2  =  4ax (opens right, a > 0)

y2  =  -4ax (opens right, a > 0)

x2  =  4ay (opens up, a > 0)

x2  =  -4ay (opens down, a > 0)

Vertex at (h, k) :  

(y - k)2  =  4a(x - h) (opens right, a > 0)

(y - k)2  =  -4a(x - h) (opens right, a > 0)

(x - h)2  =  4a(y - k) (opens up, a > 0)

(x - h)2  =  -4a(y - k) (opens down, a > 0)

Equation of a Parabola in Vertex form

Vertex at Origin :  

y  =  ax2 (opens up, a > 0)

y  =  -ax2 (opens down, a > 0)

x  =  ay2 (opens right, a > 0)

x  =  -ay2 (opens left, a > 0)

Vertex at (h, k) :  

y  =  a(x - h)2 + k (opens up, a > 0)

y  =  -a(x - h)2 + k (opens down, a > 0)

x  =  a(y - k)2 + h (opens right, a > 0)

y  =  -a(y - k)2 + h (opens left, a > 0)

Step-by-step explanation:

7 0
3 years ago
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