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kotegsom [21]
3 years ago
5

Also please let me know if the first 3 are incorrect

Mathematics
1 answer:
zhuklara [117]3 years ago
4 0
They look good keep up the good work
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Plls help im failing
allsm [11]

Answer:

9^35

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
PLZ HELP ASAP WILL MARK BRAINLIEST
Mkey [24]

DEF will be 2.68

EF will be .67

3 0
3 years ago
Which operation should you perform last in the expression 12 × 4^2?
drek231 [11]

Answer:

Multiplication

Step-by-step explanation:

Follow GEMDAS or PEMDAS. I would recommend GEMDAS. GEMDAS stands for Grouping symbols, exponents, multiply, divide, addition, subtraction. GO form the order G to S

3 0
3 years ago
Which function has the following characteristics?
iren [92.7K]

Using the asymptote concept, the function with a vertical asymptote at x = 3 and an horizontal asymptote at y = -\frac{1}{2} is given by:

f(x) = -\frac{x}{2(x - 3)}

<h3>What are the asymptotes of a function f(x)?</h3>

  • The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.
  • The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity.

The vertical asymptote at x = 3 means that x = 3 is a root of the denominator, hence:

f(x) = \frac{g(x)}{x - 3}

The horizontal asymptote at y = -1/2 means that:

\lim_{x \rightarrow \infty} f(x) = -\frac{1}{2}

Which happens if g(x) = -\frac{x}{2}, hence the function is:

f(x) = -\frac{x}{2(x - 3)}

More can be learned about asymptotes at brainly.com/question/16948935

#SPJ1

4 0
2 years ago
Let $DEF$ be an equilateral triangle with side length $3.$ At random, a point $G$ is chosen inside the triangle. Compute the pro
umka21 [38]

|\Omega|=(\text{the area of the triangle})=\dfrac{a^2\sqrt3}{4}=\dfrac{3^2\sqrt3}{4}=\dfrac{9\sqrt3}{4}\\|A|=(\text{the area of the sector})=\dfrac{\alpha\pi r^2}{360}=\dfrac{60\pi \cdot 1^2}{360}=\dfrac{\pi}{6}\\\\\\P(A)=\dfrac{\dfrac{\pi}{6}}{\dfrac{9\sqrt3}{4}}\\\\P(A)=\dfrac{\pi}{6}\cdot\dfrac{4}{9\sqrt3}\\\\P(A)=\dfrac{2\pi}{27\sqrt3}\\\\P(A)=\dfrac{2\pi\sqrt3}{27\cdot3}\\\\P(A)=\dfrac{2\pi\sqrt3}{81}\approx13.4\%

8 0
3 years ago
Read 2 more answers
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