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3 years ago

15

Susie watched 1/4

1 answer:

Annette [7]3 years ago

8 0

Answer2 since she watched 1/4 so she watched 1 then at night so 2

Step-by-step explanation:

You might be interested in

What is the average rate of change of f(x), represented by the table of values, over the interval [-3,4]

77julia77 [94]

Answer:

Average rate of change = 3

Step-by-step explanation:

We know that the average rate of change is basically a measure of the slope of the secant line in the closed interval [a, b].

here a = -3 and b = 4

f(b) = f(4) = 27

f(a) = f(-3) = 6

⇒ Average rate of change = f(b) - f(a) / b - a

= (27 - 6) / (4 - (-3))

= 21 / 7

= 3

Therefore, Average rate of change = 3

6 0

3 years ago

Find the first four terms of the sequence an = 4n-6

natulia [17]

Answer:

The first four terms of the sequence are-6, -2, 2 and 6

Step-by-step explanation:

The given sequence is $a_n=4n-6$

In order to find the first four term of the sequence we put n =0, 1, 2 and 3.

For n =0

$a_0=4(0)-6=-6$

For n =1

$a_1=4(1)-6=-2$

For n =2

$a_2=4(2)-6=2$

For n =3

$a_3=4(3)-6=6$

Therefore, the first four terms of the sequence are

-6, -2, 2 and 6

5 0

3 years ago

-1/225x^2+2/3x what is the vertex of this quadratic function?

Julli [10]

$\bf \textit{vertex of a parabola}\\ \quad \\\\

\begin{array}{lccclll}
y=&-\frac{1}{225}x^2&+\frac{2}{3}x\\\\
y=&-\frac{1}{225}x^2&+\frac{2}{3}x&+0\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

6 0

3 years ago

If you take a number, times by 7 then subtract 1. You get the same as if you took the number, times by 5 then subtract 8. What i

valentinak56 [21]

Answer:

1 i think

Step-by-step explanation:

6 0

3 years ago

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

3 years ago