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2 years ago

How do you solve simple linear equations

Mathematics

2 answers:

lesya692 [45]2 years ago

8 0

Answer:

See Explanation

Step-by-step explanation:

I think you mean how do you find linear equations. It is actually very simple when you memorize the formula. Linear equations are in the form of y = mx + b. m is the slope. b is the y-intercept.

ex. given: slope = 5, y-intercept = -3: our equation is y = 5x - 3

Slope is rise/run

y-intercept is (0,y) or you substitute.

Vera_Pavlovna [14]2 years ago

4 0

It’s easy just make a line and put numbers each line to solve the linear equation and than solve it simple

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1. If a scale factor is applied to a figure and all dimensions are changed proportionally, what is the effect on the perimeter o

tester [92]

Answer:

Part 1) The perimeter of the new figure must be equal to the perimeter of the original figure multiplied by the scale factor (see the explanation)

Part 2) The area of the new figure must be equal to the area of the original figure multiplied by the scale factor squared

Part 3) The new figure and the original figure are not similar figures (see the explanation)

Step-by-step explanation:

Part 1) If a scale factor is applied to a figure and all dimensions are changed proportionally, what is the effect on the perimeter of the figure?

we know that

If all dimensions are changed proportionally, then the new figure and the original figure are similar

When two figures are similar, the ratio of its perimeters is equal to the scale factor

The perimeter of the new figure must be equal to the perimeter of the original figure multiplied by the scale factor

Part 2) If a scale factor is applied to a figure and all dimensions are changed proportionally, what is the effect on the area of the figure?

we know that

If all dimensions are changed proportionally, then the new figure and the original figure are similar

When two figures are similar, the ratio of its areas is equal to the scale factor squared

The area of the new figure must be equal to the area of the original figure multiplied by the scale factor squared

Part 3) What would happen to the perimeter and area of a figure if the dimensions were changed NON-proportionally? For example, if the length of a rectangle was tripled, but the width did not change? Or if the length was tripled and the width was decreased by a factor of 1/4?

we know that

If the dimensions were changed NON-proportionally, then the ratio of the corresponding sides of the new figure and the original figure are not proportional

That means

The new figure and the original figure are not similar figures

therefore

Corresponding sides are not proportional and corresponding angles are not congruent

A) If the length of a rectangle was tripled, but the width did not change?

Perimeter

The original perimeter is P=2L+2W

The new perimeter would be P=2(3L)+2W ----> P=6L+2W

The perimeter of the new figure is greater than the perimeter of the original figure but are not proportionals

Area

The original area is A=LW

The new area would be A=(3L)(W) ----> A=3LW

The area of the new figure is three times the area of the original figure but its ratio is not equal to the scale factor squared, because there is no single scale factor

B) If the length was tripled and the width was decreased by a factor of 1/4?

Perimeter

The original perimeter is P=2L+2W

The new perimeter would be P=2(3L)+2(W/4) ----> P=6L+W/2

The perimeter of the new figure and the perimeter of the original figure are not proportionals

Area

The original area is A=LW

The new area would be A=(3L)(W/4) ----> A=(3/4)LW

The area of the new figure is three-fourth times the area of the original figure but its ratio is not equal to the scale factor squared, because there is no single scale factor

5 0

2 years ago

How many solutions does this equation have 3(x+1)-5=3x-2

4vir4ik [10]

Answer:

Infinite.

Step-by-step explanation:

3(x+1)-5=3x-2

3x+3-5=3x-2

3x-2=3x-2

True for any number.

6 0

2 years ago

Read 2 more answers

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:

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