3 years ago

Help needed ASAP plz plz

Mathematics

1 answer:

s2008m [1.1K]3 years ago

6 0

The answer is A, -2/3

You might be interested in

Ming drew the model below to represent the equation 24 + 12 = blank x (8 + 4).

MrMuchimi

Answer:

Step-by-step explanation:

24+12 = 36

8+4 = 12

36 = _ x 12

_ is 3 since 3x12 = 36

6 0

3 years ago

Read 2 more answers

Which pair of angles are vertical angles? 6 and 5 2 and 3 1 and 3 5 and 8

dimulka [17.4K]

Answer:

1 and 3

Step-by-step explanation:

1 and 3 are vertical angles

7 0

2 years ago

Determine the end behavior for function f(x)=-x^4+5x^3-3

Gnesinka [82]

Answer:

Step-by-step explanation:

The dominant term of this function is x^4. The graph of x^4 starts in Quadrant II and continues in Quadrant I.

If we have y = -x^4, the graph starts in Quadrant III and continues in Quadrant IV. This is the end behavior for f(x)=-x^4+5x^3-3.

4 0

3 years ago

The graph shows a probability distribution P(X)P(X) for a discrete random variable X.

andrey2020 [161]

So what are you asking lmao

5 0

3 years ago

Integration of sin^2 2x cos^3 2x dx

lilavasa [31]

$\int\left[\sin^2(2x)\cos^3(2x)\right]dx=\int\left[\sin^2(2x)\cos^2(2x)\cos(2x)\right]dx\\\\\text{Use the trigonometric identity}\sin^2(x)+\cos^2(x)=1\to\cos^2(x)=1-\sin^2(x)\\\\=\int\left\{\sin^2(2x)[1-\sin^2(2x)]\cos(2x)\right\}dx\\\\\text{substitute}\ \sin2x=t\ \to\ 2\cos2x\ dx=dt\ \to\ \cos2x\ dx=\dfrac{1}{2}\ dt\\\\=\dfrac{1}{2}\int\left[t^2(1-t^2)\right]dt=\dfrac{1}{2}\int(t^2-t^4)dt=\dfrac{1}{2}\left(\dfrac{1}{3}t^3-\dfrac{1}{5}t^5\right)+C$

$=\dfrac{1}{2}\left(\dfrac{1}{3}\sin^3(2x)-\dfrac{1}{5}\sin^5(2x)\right)+C$

4 0

3 years ago