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gizmo_the_mogwai [7]
3 years ago
14

How much is +17 + -17 + -43=

Mathematics
2 answers:
snow_tiger [21]3 years ago
6 0
-43 i believe is the answer
Novosadov [1.4K]3 years ago
6 0
Negative 43 because a -17 + 17 cancels each other out :) hope this helped!
You might be interested in
F= 3x^2 - x +4 , find f (1/2)
sergey [27]
The answer is 7 i did it on a calculator lol
4 0
3 years ago
<img src="https://tex.z-dn.net/?f=F%28x%29%20%3D%20%5Cfrac%7Bx%5E%7B3-8%7D%20%7D%7Bx%5E%7B2%7D%20-6x%2B8%7D" id="TexFormula1" ti
Firlakuza [10]

Answer:

  • Domain: All the real values except x = 2 and x = 4: R - {2, 4}
  • Holes: x = 2
  • VA, vertical asymptores: x = 4
  • HA: horizontal asymptotes: there are not horizontal asymptotes
  • OA: oblique asymptotes: x + 6 [note that OH does not stand for any known feature, and so it is understood that it was intended to write OA]
  • Roots: x = 2
  • Y-intercept: -1

Step-by-step explanation:

1. <u>Given</u>:

f(x)=\frac{x^3-8}{x^2-6x+8}

  • Note that the number 8 in the numerator is not part of the power.
  • Type of function: rational function

2. <u>Domain</u>: is the set of x-values for which the function is defined.

The given function is defined for all x except those for which the denominator equals 0.

  • Denominator:  x² -6x + 8 = 0
  • Solve for x:

        Factor. (x - 4 )(x - 2) = 0

        Zero product property: (x - 4) = 0 or (x - 2) = 0

        x - 4 = 0 ⇒ x = 4

        x - 2 = 0 ⇒ x = 2

  • Domain:

        All the real values except x = 2 and x = 4: x ∈ R / x ≠ 2 and x ≠ 4.

3. <u>Holes</u>:

The holes on the graph of a rational function are at those x-values for which both the numerator and denominator are zero.

  • Find the values for which the numerator is zero:

        Numerator: x³ - 8 = 0

        Factor using difference of cubes property:

                   a³ - b³ = (a - b)(a² + ab + b²)

                   x³ - 8 = (x - 2)(x² + 2x + 4) = 0

        Zero product property:  (x - 2)(x² + 2x + 4) = 0

                    x - 2 = 0 ⇒ x = 2                    

                    x² + 2x + 4 = 0 (this has not real solution)

  • The values for which the denominator is zero were determined above: x = 2 and x = 4.

  • Conclusion: for x = 2 both numerator and denominator equal 0, so this is a hole.

4. <u>VA: Vertical asymptotes</u>.

The vertical asymptotes on the graph of a rational function are the vertical lines for which only the denominator (and not the numerator) equals zero.

  • In the previous part it was determined that happens when x = 4.

5. <u>HA: Horizontal asymptotes</u>.

In rational functions, if the numerator is a higher degree polynomial than the denominator, there is no horizontal asymptote.

6. <u>OA: oblique asymptotes</u>

  • Find the quotient and the remainder.

                       x + 6

                  _______________

x² - 6x + 8 )   x³ + 0x² + 0x - 8

                  - x³ + 6x² - 8x

                   ___________

                          6 x² -   8x -  8

                        - 6x² + 36x - 48

                        _____________

                                    28x  - 56

Result: (x + 6) + (28x - 56) / (x² - 6x + 8)

  • Find limit x → ∞

\lim_{x \to \infty}(x + 6) + \frac{28x-56}{x^2-6x+8}=x+6

<u>7. Roots</u>:

Roots are the values for which f(x) = 0.

That happens when the numerator equals 0, and the denominator is not 0.

As determined earlier: x³ - 8 = 0 ⇒ x = 2.

8. <u>Y-Intercept</u>

The y-intercepts of any function are the y-values when x = 0

  • Substitute x = 0 into the function:

         f(x)=\frac{x^3-8}{x^2-6x+8}=\frac{0^3-8}{0^2-6(0)+8}}=\frac{-8}{8} =-1

3 0
4 years ago
Please help me( '人' )​
love history [14]

Answer:

S(1,2)

Step-by-step explanation:

The value that is being altered is the y value if you go from Q to T and then T to S. The x value = 1 and remains the same for point S.

To go from Q to T, you go from 8 to 5 on the y value. Remember x remains the same. That's three units (8 - 5 = 3)

To go from T to S must be 3 units as well, since the small diagonal is bisected. 5 - 3 = 2 is the answer.

So point S is noted as S(1,2)

4 0
2 years ago
A number m is no less than –1 or less than or equal to −5.
Law Incorporation [45]

Answer:

-1 < m ≤ - 5

Step-by-step explanation:

Your expression includes: m, -1, -5, >, ≤

m is the value being compared so it will be in the middle.

# sign m sign #

m is no less than -1: no less than means it is greater than -1, but does not include -1. Therefore, our sign is >

m > -1

m is less than or equal to -5. This means we use ≤ because it can equal -5.

m ≤ -5

Now combine the two expressions

m > -1 – This can also be written as -1 < m.

m ≤ -5

-1 < m ≤ - 5

7 0
3 years ago
Which equation demonstrates the additive identity property?
Bezzdna [24]

Answer:

Step-by-step explanation:

Solve for o in the equation (7 + 4) + (7 - 41) = 14©(7 + 4) + 0 = 7 + 41o(7 + 4)(1) = 7 + 41o(7 + 41) + ( - 7 - 41) = 0

We first need to simplify the expression removing parentheses

Simplify 41o(7 + 4): Distribute the 41o to each term in (7+4)

41o * 7 = (41 * 7)o = 287o 41o * 4 = (41 * 4)o = 164o

Our Total expanded term is 287o + 164o

Simplify 41o(7 + 41): Distribute the 41o to each term in (7+41)

41o * 7 = (41 * 7)o = 287o 41o * 41 = (41 * 41)o = 1681o

Our Total expanded term is 287o + 1681o

Our updated term to work with is (7 + 4) + (7 - 41) = 14©(7 + 4) + 0 = 7 + 287o + 164o(1) = 7 + 287o + 1681o + ( - 7 - 41) = 0

We first need to simplify the expression removing parentheses

Simplify 164o(1): Distribute the 164o to each term in (1)

164o * 1 = (164 * 1)o = 164o Our Total expanded term is 164o

Our updated term to work with is (7 + 4) + (7 - 41) = 14©(7 + 4) + 0 = 7 + 287o + 164o = 7 + 287o + 1681o + ( - 7 - 41) = 0

Step 1: Group variables: We need to group our variables (7 and 14©(7. To do that, we subtract 14©(7 from both sides (7 - 14©(7 = 14©(7 - 14©(7

Step 2: Cancel 14©(7 on the right side: 0o = 0 Step 3: Divide each side of the equation by 0

0o 0 = 0 0 o =

5 0
3 years ago
Read 2 more answers
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