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Natasha_Volkova [10]
3 years ago
14

Ab=45 mm,bc=10cm 4mm ,cd = 5 cm 6 mm ,de= 35 mm,ea=x cm, and the perimeter of irregular pentagon is 280 mm. Find the length of t

he unknown side x in cm
Mathematics
1 answer:
DaniilM [7]3 years ago
3 0

Answer:

The length of the unknown side x is 4 cm

Step-by-step explanation:

we know that

The perimeter of a polygon is equal to the sum of its length sides

Remember that

1 cm=10 mm

In this problem we have

AB=45 mm ------> convert to cm -----> AB=45/10=4.5 cm

BC=10 cm 4 mm -----> convert to cm -----> BC=10+4/10=10.4 cm

CD=5 cm 6 mm  -----> convert to cm -----> CD=5+6/10=5.6 cm

DE=35 mm ------> convert to cm -----> DE=35/10=3.5 cm

EA=x cm

P=280 mm ------> convert to cm -----> P=280/10=28 cm

The perimeter of the irregular polygon is equal to

P=AB+BC+CD+DE+EA

substitute the values

28=4.5+10.4+5.6+3.5+x

28=24+x

x=28-24

x=4 cm

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Answer:

9.20rs

Step-by-step explanation:

4.60*2=9.20

hope this answer helps

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3 years ago
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7(6+d)=49 Solve for d.<br><br> A:d=7<br> B:d=-1<br> C:d=-7<br> D:d=1
Elan Coil [88]

Answer: d = 1

Step-by-step explanation:

First distribute the 7 to each term of (6 + d) = 42 + 7d = 49 now subtract 42 from both sides which = 7d = 7. Divide by 7 = 7d/7 = 7/7 = d = 1

3 0
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In a history class, the girl to boy ratio is 5 to 3. If there are 80 students total, how many boys are there?​
Brilliant_brown [7]
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50/30

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8 0
2 years ago
If a quadratic equation has 4-i as a solution, what must the other solution be?
Luden [163]

Answer:

4+i

Step-by-step explanation:

A complex number usually took the form a+bi where a and b are real numbers and 'i' represents an imaginary number. For a quadratic equation, the complex roots for the root of a quadratic equation took the form known as complex conjugates. The complex conjugates are formed by changing the sign of the imaginary part.

SO, if a quadratic equation has 4-i as a solution, the other solution must be 4+i.

6 0
2 years ago
NEED HELP Find values for a, b, c, and d so that the following matrix product equals the 2X2 identity matrix. Explain or show ho
ehidna [41]

We want to find the values of a, b, c, and d such that the given matrix product is equal to a 2x2 identity matrix. We will solve a system of equations to find:

  • b = 1
  • a = -1
  • c = -2
  • d = -1

<h3>Presenting the equation:</h3>

Basically, we want to solve:

\left[\begin{array}{cc}-1&2\\a&1\end{array}\right]*\left[\begin{array}{cc}b&c\\1&d\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]

The matrix product will be:

\left[\begin{array}{cc}-b + 2&-c + 2d\\a*b + 1&a*c + d\end{array}\right]

Then we must have:

-b + 2 = 1

This means that:

b = 2 - 1 = 1

  • b = 1

We also need to have:

a*b + 1 = 0

we know the value of b, so we just have:

a*1 + b = 0

  • a = -1

Now the two remaining equations are:

-c + 2d = 0

a*c + d = 1

Replacing the value of a we get:

-c + 2d = 0

-c + d = 1

Isolating c in the first equation we get:

c = 2d

Replacing that in the other equation we get:

-(2d) + d = 1

-d = 1

  • d = -1

Then:

c  = 2d = 2*(-1) = -2

  • c = -2

So the values are:

  • b = 1
  • a = -1
  • c = -2
  • d = -1

If you want to learn more about systems of equations, you can read:

brainly.com/question/13729904

4 0
3 years ago
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