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Digiron [165]
3 years ago
12

Solve the equation 15x + 8 = y + 13 for x if y is 40. Make sure to first solve the equation for x in terms of y.

Mathematics
2 answers:
Nataly_w [17]3 years ago
4 0

Answer:

3

Step-by-step explanation:

15x +8 =40 +13

15x + 8=53

15x=45

3

timama [110]3 years ago
4 0

Answer:

x=3

Step-by-step explanation:

15x+8=(40)+13

15x+8=53

subtract 8 from both sides to isolate the variable.

15x=45

divide both sides by 15

x=3

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Find the measure of each exterior angle of a regular 56-gon.
Karo-lina-s [1.5K]
The answer is 6 degree.

Step by step process

The formula for calculating the measure of each exterior angle of a polygon is (360/n).

The number of sides of 56-gon(n) = 56.


So, each exterior angle
= (360/56)
= 6.43 or
6(approximately)
4 0
3 years ago
Let a be a rational number and b be an irrational number. Which of the following are true statements?(there is more than 1 answe
kramer
<span>A.) the sum of a and b is never rational.
This is a true statement. Since an irrational umber has a decimal part that is infinite and non-periodical, when you add a rational number to an irrational number, the result will have the same infinite non periodical decimal part, so the new number will be irrational as well.

</span><span>B.) The product of a and b is rational
This one is false. Zero is a rational number, and when you multiply an irrational number by zero, the result is always zero.

</span><span>C.) b^2 is sometimes rational
This one is true. When you square an irrational number that comes from a square root like </span>\sqrt{2}, you will end with a rational number: ( \sqrt{2} )^{2}=2, but, if you square rationals from different roots than square root like \sqrt[3]{2}, you will end with an irrational number: \sqrt[3]{2^{2} } = \sqrt[3]{2}. 

<span>D.) a^2 is always rational
This one is false. If you square a rational number, you will always end with another rational number.

</span><span>E.) square root of a is never rational
</span>This one is false. The square root of perfect squares are always rational numbers: \sqrt{64} =8, \sqrt{16} =4,...

F.) square root of b is never rational
This one is true. Since the square root of any non-perfect square number is irrational, and all the irrational numbers are non-perfect squares, the square root of an irrational number is always irrational.

We can conclude that given that<span> a is a rational number and b be an irrational number, A, C, D, and F are true statements.</span>
4 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
2 years ago
1/3n-7/12=-3/8 n=5/72 n=-5/72 n=5/8 n=-2 7/8
prisoha [69]

Given the equation;

\frac{1}{3}n-\frac{7}{12}=-\frac{3}{8}

To solve, let us multiply through by the least common multiple of the the denominator of the three fractions wich is 24;

3 0
1 year ago
Help asap idek what to do
daser333 [38]

So to solve this question, your goal is to find out how the way it is solved is not correct.

Your answer would be: On the third line, the student adds the 8 to both sides instead of subtracting. The way the initial equation is given is

y-(-8)=-6(x-2). After distributing the six, the student should make the 8 positive because subtracting a negative makes a positive. After solving, the equation should look like: y(+8)=-6x+12, so you would subtract the 8 from both sides instead of adding it, and solve from there.

5 0
3 years ago
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