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44.

We have that    y  = Acos (Bθ)

If the max is 4  and the minimun  is -4, the amplitude   =  4  =  A

To find  B, we can solve this

2pi / (2pi / 3)  = B

3 * 2pi  / 2pi  = B

3  = B

So....the function is

y   = 4 cos 3θ

45.  tan  3pi/4   =   - tan pi/4  =    -1

42.

s  = r  * (theta in radians)   ...so we have

s   = 3  *  (5pi/3)    =  5pi ≈  15.7 in

43.   Here's the graph....

https://www.desmos.com/calculator/9btjsjrpqx

40.

Amplitude  =  l  -5  l    =  5

Period....we can solve this......   2pi  / 6  = period   = (1/3) pi    

Last answer

41.  cos 315°  is a 45° reference angle in the 4th quadrant....and the cosine is positive in this quadrant.. so..

  cos 45°   =   √2 / 2

If he named 1 number, then the last number he named would be 15

If he named 2 numbers, the last number he named would be 15 + 6

If he named 3 numbers, the last number he named would be 15 + 6 + 6

If he named 4 numbers, the last number he named would be 15 + 6 + 6 + 6  which is  15 + 6(4 - 1)

So... if he named  n  numbers, the last number he named would be  15 + 6(n - 1)

They tell us the last number he named was  363 . So..

363  =  15 + 6(n - 1)    Subtract  15  from both sides.

348  =  6(n - 1)            Divide both sides by  6 .

58  =  n - 1                  Add  1  to both sides.

59  =  n

[L - F] / D + 1=N, where L=Last term, F=First term, D=Common difference, N=Number of terms

[363 - 15] / 6 + 1 =59 terms or numbers that Oliver counted.

Its zero

0⁰  is an indeterminant form...here's why ...

Note...if we let n be any real number...we have

0⁰  =  0 ^n-n  =  0ⁿ  / 0ⁿ   =    0  / 0

But also note that    0  / 0   =  pi        or  0 / 0    =  11   or   0 / 0  = e  ......  etc....

Thus....we  can't actually determine  the value  of  0^n-n  = 0⁰  

hi beary!

if you ask the calculator, it says it is zero.

however, the alternating sum of binomial coefficients from the n-th row of Pascal's triangle is what you obtain by expanding (1-1)^n using the binomial theorem, i.e., 0n.

 But the alternating sum of the entries of every row except the top row is 0, since 0^k=0 for all k greater than 1.

But the top row of Pascal's triangle contains a single 1, so its alternating sum is 1, which supports the notion that (1-1)^0=0^0 if it were defined, should be 1.

so therefore 0^0 is zero.

i hope this helped,

gavin

Yep your correct on both.

SA for Pyramid 

2*l*/+l^2

l=length of base

/=slant height

Volume for Pyramid 

1/3*B*h

B is area of base

SA for cone

3.14*r*/+3.14*r^2

/ equals slant height

Volume for cone

1/3*3.14*r^2*h

h is the height

Hope this helps!

Donut🍩🍩🍩

Need help asap!

1. At a certain time of day, a tree has a shadow 50 feet long. At the same time, a 6-foot-tall man stands next to the tree, and his shadow is 20 feet long. What is the height of the tree, in feet?

You are very welcome :)

You are correct on both questions. 

2. The terms $140, a, \frac{45}{28}$ are the first, second and third terms, respectively, of a geometric sequence. If $a$ is positive, what is the value of $a$?

\(\begin{array}{|rcll|} \hline a &=& \sqrt{140\cdot \frac{45}{28}} \\ a &=& 15 \\ \hline \end{array}\)

"2. The terms $140, a, \frac{45}{28}$ are the first, second and third terms, respectively, of a geometric sequence. If $a$ is positive, what is the value of $a$?"

"1. Consider the infinite arithmetic sequence $A$ with first term $5$ and common difference $-2$. Now define the infinite sequence $B$ so that the $k^{th}$ term of $B$ is $2$ raised to the $k^{th}$ term of $A$. Find the sum of all of the terms of $B$."

.

Let $C$ and $D$ be points on the semicircle with diameter $\overline{AB}$. If $AD = BC = 3$ and $CD = 7$,

then find the radius of the semicircle.

1. sin rule:

\(\begin{array}{|rcll|} \hline \dfrac{ \sin(180^{\circ}-2\varphi) } {7} &=& \dfrac{ \sin(\varphi) }{r} \\ \dfrac{2\sin(\varphi)\cos(\varphi)}{7} &=& \dfrac{\sin(\varphi)}{r} \\ \dfrac{2 \cos(\varphi)}{7} &=& \dfrac{1}{r} \\ \cos(\varphi) &=& \dfrac{7}{2r} \\ \hline \end{array} \)

2. cos rule:

\(\begin{array}{|rcll|} \hline 3^2 &=& r^2+r^2-2rr\cos(\varphi) \\ 9 &=& 2r^2(1-\cos(\varphi)) \\ \dfrac{9}{2r^2} &=& 1-\cos(\varphi) \quad & | \quad \cos(\varphi) = \dfrac{7}{2r} \\ \dfrac{9}{2r^2} &=& 1-\dfrac{7}{2r}\quad & | \quad \cdot 2r^2 \\ 9 &=& 2r^2 -7r \\ 2r^2 -7r -9 &=& 0 \\ r &=& \dfrac{7 \pm \sqrt{49-4\cdot 2\cdot (-9) } } {2\cdot 2} \\ &=& \dfrac{7 \pm \sqrt{49+72 } } {4} \\ &=& \dfrac{7 \pm \sqrt{121 } } {4} \\ r &=& \dfrac{7 \pm 11 } {4} \quad & | \quad r > 0\ !\\ &=& \dfrac{7 +11 } {4} \\ &=& \dfrac{18 } {4} \\ \mathbf{r} &\mathbf{=}& \mathbf{4.5} \\ \hline \end{array}\)

The radius of the semicircle is 4.5

40)

You are given two triangles and the information that the three pairs of corresponding angles are congruent. What other information would guarantee that the triangles are congruent?

D: One pair of corresponding sides has the same measure.

43)

If minor arc AC = 96°, what is the measure of ∠ABC?

Here's how I have to think of these....

Draw the angle that forms minor arc AC...its measure is 96° .

Even though it wasn't specifically stated in the question, I assume that lines AB and BC are tangent to the circle. That means line AB and line BC each forms a right angle with the line drawn from the center of the circle to point A and point C respectively.

The sum of the angle measures in every quadrilateral is  360° . So....

m∠ABC  =  360° - 90° - 90° - 96°  =  180° - 96°  =  84°

32)

Since the sum of the angle measures in every triangle is  180°,

and we know that two of the angles are  90°  and  45°.....

the measure of the third angle   =   m∠A   =   180° - 90° - 45°   =   45°

Since  m∠A  =  m∠C ,  triangle  ABC  is isosceles and   AB  =  BC

By the Pythagorean Theorem...

(AB)² + (BC)²  =  n²

                                      We can replace  AB  with  BC  since they are the same length.

(BC)² + (BC)²  =  n²

                                      Combine like terms.

2(BC)²  =  n²

                                      Divide both sides of the equation by  2 .

(BC)²  =  n² / 2

                                      Take the positive square root of both sides.

BC  =  \(\sqrt{\frac{n^2}{2}}\)

                                      Simplify.

BC  =  \(\frac{\sqrt{n^2}}{\sqrt2}\)

BC  =  \(\frac{n}{\sqrt2}\)

                                      Rationalize denominator.

BC  =  \(\frac{n\sqrt2}{2}\)

Whats the answer?

Thanks so much!!!

2x^2 + 15x  + 25

Here's a method that often works for problems like these...it's a little guesswork....but sometimes it pays off

I'm guessing that we can   write  15x   in this  manner......  10x  +  5x

2x^2 + 10x  + 5x  +  25

I see that  the first constant multiplied by 5 will give me the second constant.....and the third constant  multiplied by 5  will give me the last constant...so......let's factor. the first two  terms and the last two  terms....   [factoring  by "grouping " ]

2x ( x + 5 )  + 5 ( x + 5).....  note that  ( x + 5)  is common

So......the factorization  is

( x + 5 )  ( 2x + 5)