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solmaris [256]
3 years ago
14

What is the area of a shape that has the lengh of 10 cm and the height of 5cm

Mathematics
1 answer:
Andreyy893 years ago
4 0
Assuming you mean a shape that is a paralellogram (paralellogram has 4 sides, has 2 pairs of parralell sides)


area=legnth times width
length=10
width=5
area=10 times 5

area is 2-D
10 times 5=50 square centimeters

the asnwer is 50 square cm
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He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0. which best
seropon [69]

The question is incomplete. Completed question is given below the answer.

His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Given  

Then the solution follows thus:

Step 1: 8(x – 4) = 2(x + 2)

Step 2: 4(x – 4) = (x + 2)

Step 3: 4x – 16 = x + 2

Step 4: 3x = 18

Step 5: x = 6

It can be seen that his solution is correct. But 6 is not an extraneous solution.

An extraneous solution is a solution to an equation that emerges from the process of solving the problem but is not a valid solution to the original problem.

When 6 is substituted into the original equation, the original equation holds.

Therefore, his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Learn more about extraneous solution here: brainly.com/question/3751209

#SPJ4

Completed question:-

A student solves the following equation for all possible values of x:His solution is as follows:

Step 1: 8(x – 4) = 2(x + 2)

Step 2: 4(x – 4) = (x + 2)

Step 3: 4x – 16 = x + 2

Step 4: 3x = 18

Step 5: x = 6

He determines that 6 is an extraneous solution because the difference of the numerators is 6, so the 6s cancel to 0.

Which best describes the reasonableness of the student’s solution?

His solution for x is correct and his explanation of the extraneous solution is reasonable.

His solution for x is correct, but in order for 6 to be an extraneous solution, both denominators have to result in 0 when 6 is substituted for x.

His solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

His solution for x is incorrect. When solved correctly, there are no extraneous solutions.

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Answer:

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Step 3: Multiply by the scalar (1 over the determinant): Aâ’1 = a11 a12 a21 a22 a11 = a12 = a21 = a22 =
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Answer:

a11 = 1

a12 = -1

a21 = -1.5

a13 = 2

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Answer:

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Step-by-step explanation:

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