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mixer [17]
3 years ago
9

Describe and correct the error(s) made in each of the problems below.

Mathematics
1 answer:
ladessa [460]3 years ago
6 0

Answer:

\displaystyle \frac{1-x}{(5-x)(-x)} =-\frac{x-1 }{ x(x-5)}

\displaystyle \frac{5}{s}\times \frac{2}{5} =\frac{2}{s}

Step-by-step explanation:

<u>Errors in Algebraic Operations </u>

It's usual that students make mistakes when misunderstanding the application of algebra's basic rules. Here we have two of them

  • When we change the signs of all the terms of a polynomial, the expression must be preceded by a negative sign
  • When multiplying negative and positive quantities, if the number of negatives is odd, the result is negative. If the number of negatives is even, the result is positive.
  • Not to confuse product of fractions with the sum of fractions. Rules are quite different

The first expression is

1-x / (5-x)(-x)=x-1 / x(x-5)

Let's arrange into format:

\displaystyle \frac{1-x}{(5-x)(-x)} =\frac{x-1 }{ x(x-5)}

We can clearly see in all of the factors in the expression the signs were changed correctly, but the result should have been preceeded with a negative sign, because it makes 3 (odd number) negatives, resulting in a negative expression. The correct form is

\displaystyle \frac{1-x}{(5-x)(-x)} =-\frac{x-1 }{ x(x-5)}

Now for the second expression

5/s+2/5=2/s

Let's arrange into format

\displaystyle \frac{5}{s}+\frac{2}{5} =\frac{2}{s}

It's a clear mistake because it was asssumed a product of fractions instead of a SUM of fractions. If the result was correct, then the expression should have been

\displaystyle \frac{5}{s}\times \frac{2}{5} =\frac{2}{s}

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0.25 in a fraction would be 1/4
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A ramp with a one meter distance from edge to edge reaches a height of 0.643 meters. What is the value of the acute angle?
natita [175]

The value of the acute angle is 40°

Step-by-step explanation:

Let the angle be 'θ'

Length of the ramp = 1 m

Height = 0.643 m

To find :

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we can use sine to find the angle of elevation.

sin θ = 0.643/1

sin θ = 0.643

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sin θ = sin 40°

So, θ = 40°

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3 years ago
Find the area of the following shape.
ad-work [718]

Answer:

57 units^2

Step-by-step explanation:

First find the area of the triangle on the left

ABC

It has a base AC which is  9 units and a height of 3 units

A = 1/2 bh = 1/2 ( 9) *3 = 27/2 = 13.5

Then find the area of the triangle on the right

DE

It has a base AC which is  6 units and a height of 1 units

A = 1/2 bh = 1/2 ( 6) *1  = 3

Then find the area of the triangle on the top

It has a base AC which is  3 units and a height of 3 units

A = 1/2 bh = 1/2 ( 3) *3  = 9/2 = 4.5

Then find the area of the rectangular region

A = lw = 6*6 = 36

Add them together

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3 years ago
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Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

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Parentheses:
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Multiply:
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