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2 years ago

5. ALGEBRA Refer to the figure in

Mathematics

1 answer:

mamaluj [8]2 years ago

5 0

The value of $x$ that satisfies all the expressions of the system of linear equations is 20.

<h3 /><h3>Procedure - Determination of a variable associated to a pair of corresponding angles</h3>

According to the figure, angles 1 and 5 are <em>corresponding</em> angles between two <em>parallel</em> lines, therefore we have the following equivalence:

$4\cdot x + 40 = 120$ (1)

After some algebraic handling we find the value of $x$ :

$4\cdot x = 80$

$x = 20$

The value of $x$ that satisfies all the expressions of the system of linear equations is 20. $\blacksquare$

<h3>Remarks</h3>

There is a missing figure for the statement, we proceed to include it below after some research.

To learn more on angles, we kindly invite to check this verified question: brainly.com/question/15767203

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May anyone please help me?

kiruha [24]

The answer will be D) 8

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3 years ago

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What happens to the minimum point of f(x) = x2 + 5x + 2, when it is changed to f(x) = 5x2 + 5x + 2?

Aleonysh [2.5K]

The answer is B) it shifts up and to the right

6 0

3 years ago

Can someone please help me with this question?!? I am so confused and I don't know how to answer it.

lbvjy [14]

9514 1404 393

Answer:

a. f(0) = 1

b. DNE (does not exist)

c. DNE

d. lim = 3

Step-by-step explanation:

The function exists at a point if it is defined there. The function is defined anywhere on the solid line and at solid dots. It is not defined at open circles. So, the function is defined everywhere except (2, 3), which has an open circle.

The open circle at (0, 4) prevents the function from being doubly-defined at x=0, since it is already defined to be 1 at x=0.

This discussion tells you ...

f(0) = 1

f(2) does not exist. There is a "hole" in the function definition there.

The function has a limit at a point if approaching from the left and approaching from the right have you approaching that same point.

Consider the point (1, 2). The graph is a solid line through that point. Approaching from values less than x=1, we get to the same point (1, 2) as when we approach from values greater than x=1.

Similarly, consider the point (2, 3). Approaching from values of x less than 2, we get to the same point (2, 3) as when we approach from x-values greater than 2. The limit at x=2 is 3. The only difference from the previous case is that the function is not actually defined to be that value there.

Now consider what happens at x=0. When we approach from the left, we approach the point (0, 4). When we approach from the right, we approach the point (0, 1). These are different points. Because they are different coming from the left and from the right, we say "the limit as x→0 does not exist."

In summary, ...

a) f(0) = 1

b) lim x → 0 does not exist

c) f(2) does not exist

d) lim x → 2 = 3

_____

<em>Additional comment</em>

The significance of the function not being defined at a point where the limit exists, (2, 3), is that <em>the function is not continuous there</em>. This kind of discontinuity is called "removable", because we could make the function continuous at x=2 by defining f(2) = 3 (that is, "filling the hole").

6 0

3 years ago

Choose whether it's always, sometimes, never

Keith_Richards [23]

Answer: An integer added to an integer is an integer, this statement is always true. A polynomial subtracted from a polynomial is a polynomial, this statement is always true. A polynomial divided by a polynomial is a polynomial, this statement is sometimes true. A polynomial multiplied by a polynomial is a polynomial, this statement is always true.

Explanation:

The closure property of integer states that the addition, subtraction and multiplication is integers is always an integer.

If $a\in Z\text{ and }b\in Z$ , then a+b\in Z.

Therefore, an integer added to an integer is an integer, this statement is always true.

A polynomial is in the form of,

$p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0$

Where $a_n,a_{n-1},...,a_1,a_0$ are constant coefficient.

When we subtract the two polynomial then the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

If a polynomial divided by a polynomial then it may or may not be a polynomial.

If the degree of numerator polynomial is higher than the degree of denominator polynomial then it may be a polynomial.

For example:

$f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2$

Then $\frac{f(x)}{g(x)}=x^2+5$ , which a polynomial.

If the degree of numerator polynomial is less than the degree of denominator polynomial then it is a rational function.

For example:

$f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2$

Then $\frac{g(x)}{f(x)}=\frac{1}{x^2+5}$ , which a not a polynomial.

Therefore, a polynomial divided by a polynomial is a polynomial, this statement is sometimes true.

As we know a polynomial is in the form of,

$p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0$

Where $a_n,a_{n-1},...,a_1,a_0$ are constant coefficient.

When we multiply the two polynomial, the degree of the resultand function is addition of degree of both polyminals and the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

3 0

3 years ago

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Help me out, please!

VMariaS [17]

Answer:

y = 2/3(x - 3)

Step-by-step explanation:

slope of the line is 2/3

we can use point (3,0)

formula: y - y = m(x - x)

y - 0 = 2/3 (x - 3)

y = 2/3(x - 3)

5 0

3 years ago