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(1)

Here,  s  is a slant height and  a  is the altitude.

We can find  s  with the Pythagorean theorem.

s² + 16²  =  65²

s²  =  65² - 16²

s²  =  3969

s  =  63

Now we can find  a  with the Pythagorean theorem.

a² + 33²  =  s²

a² + 33²  =  63²

a²  =  63² - 33²

a²  =  2880

a  =  √[ 2880 ]

a  =  24√5

*****edit*****

My original answer for lateral area wasn't right because I didn't take into account that the faces aren't all the same.

The same way we found  s ,  we can find the length of the other slant height.

other slant height  =  √[ 65² - 33² ]  =  56

lateral area   =   2 * (1/2) * 32 * 63   +   2 * (1/2) * 66 * 56   =   32 * 63  +  66 * 56   =   5712  square units

In the sequence

1, 2, 2, 4, 8, 32, 256,...,

each term starting from the third term) is the product of the two terms before it. For example, the seventh term is 256, which is the product of the fifth term (8) and the sixth term (32). 

This sequence can be continued forever, though the numbers very quickly grow enormous!

(For example, the 14th term is close to some estimates of the number of particles in the observable universe.)

What is the last digit of the 35th term of the sequence?

Again, the standard form of a line is  Ax + By = C  where A is a positive integer, and B and C are integers.

Using the point  (-2, -3)  and the slope  \(\frac12\) ,  the equation of the line in point-slope form is:

y - -3  =  \(\frac12\)(x - -2)

y + 3  =  \(\frac12\)(x + 2)

                                   Multiply both sides of the equation by  2 .

2(y + 3)  =  1(x + 2)

                                   Distribute.

2y + 6  =  x + 2

                                   Subtract  2y  from both sides.

6  =  x - 2y + 2

                                   Subtract  2  from both sides.

4  =  x - 2y

x - 2y  =  4

\(10g-28g+16 = -18g+16 = -2(9g+8)\)

\(\text{Did you maybe mean $10g^2 - 28 g+16$?}\)

\(\text{If so}\\ 10g^2-28g+16 = \\ 2(5g^2-14g+8) = \\ 2(g-2)(5g-4) \)

\(xy=9 \Rightarrow x \neq 0\\ y=\dfrac 9 x\\ x+y = x + \dfrac 9 x \in \mathbb{R}-\{0\}\)

We're given a point and a slope so we can easily write the equation of the line in point-slope form.

The equation of the line in point-slope form is....

y - -4  =  -\(\frac38\)(x - 5)

Now we just need to get this equation into standard form.

The standard form of a line is  Ax + By = C  where A is a positive integer, and B and C are integers.

y - -4  =  -\(\frac38\)(x - 5)

y + 4  =  -\(\frac38\)(x - 5)

                                     Multiply both sides of the equation by  8 .

8(y + 4)  =  -3(x - 5)

                                     Distribute the  8  and the  -3

8y + 32  =  -3x + 15

                                     Add  3x  to both sides of the equation.

3x + 8y + 32  =  15

                                     Subtract  32  from both sides.

3x + 8y  =  -17

The line appears to cross the y-axis at the point  (0, -1)  , so the y-intercept is  -1 .

The points  (1, 0)  and  (0, -1)  appear to be on the line. Using these two points we can find the slope.

slope  =  \(\frac{\text{rise}}{\text{run}}\ =\ \frac{y_2-y_1}{x_2-x_1}\ =\ \frac{-1-0}{0-1}\ =\ \frac{-1}{-1}\ =\ 1\)

Now we know the y-intercept is  -1  and the slope is  1. So the equation of this line in slope-intercept form is....

y  =  1x - 1

y  =  x - 1

Also....

We can see that the slope is  1  by counting the units on the graph. To get from the point  (0, -1)  to the point  (1, 0)

we have to go up 1 unit and to the right 1 unit, so the slope  =  1/1  =  1

Because it's  \(32!/((32-5)!*5!)\) you need a () at the denominator and the practice multiplier * between ! and 5!

There are 9 choose 4 = 126 ways to choose the first team.

For the next team, there are 5 remaining people to choose from so there are 5 choose 3 = 10 ways to make it.

The remaining 2 people will be the 2 person team.

126*10= \(\boxed{1260}\) ways to choose the teams.

The top right corner is Mack's goal. Bottom left is where he starts and the red spot is the spider. Blue numbers are the number of ways Mack could get to the point.

He starts at (0,0) and there is only one way to get there. There is also only one way for him to get to all the points on the bottom and left edges, because he can only move up and right.

To get to (1,1), there are 2 ways to get there because he could come from (0,1) or (1,0).

To get to (2,1) he could come from (2,0) or (1,1). The number of ways he could get to (2,1) is the sum of the ways he could get there from the point below or the point to the left so 2+1=3 ways to get to (2,1).

If you look at the drawing, all the blue numbers are the sum of the blue numbers 1 below and 1 to the left. You might notice that this looks like Pascal's triangle! (Because it pretty much is).

Continuing to sum the numbers in this manner, the number of ways to get to (5,7) is \(\boxed{426}\)

There are other ways to solve this problem using combinations, and if this is a homework problem that is probabally the way you know how to solve it. I like this way better .

\(a=4~m/s^2\\~\\ s_0 = 0\\ s(t) = \dfrac a 2 t^2 + v_0 t = 2t^2 + v_0 t\\ s(6)-s(5)=38~m\\ 2(6^2)+6v_0 - 2(5^2)-5v_0 = 38\\ (72-50)+v_0= 38\\ v_0 = 38-22 = 16~m/s\)

If we wanted to use the average speed

\(v(t) = v_0 + at\\ \bar{v}=\dfrac{v(6)+v(5)}{2} = \dfrac{11a+2v_0}{2}= \dfrac{11}{2}a+v_0=22+v_0\\ (22+v_0)~m/s\cdot 1~s = 38~m\\ v_0 = 38-22 = 16~m/s\)

Find all real numbers \(t\) such that \(\frac23 t - 1 \ <\ t + 7\ \le\ -2t + 15\) . Give your answer as an interval.

_________

\(\frac23 t - 1 \ <\ t + 7\\~\\ \frac23 t - t \ <\ 1 + 7\\~\\ -\frac13t\ <\ 8\\~\\ t\ >\ -24\)

And

\(t + 7\ \le\ -2t + 15\\~\\ t+2t\ \leq\ -7+15\\~\\ 3t\ \leq\ 8\\~\\ t\leq\frac83\)

t  is bigger than  -24  but less than or equal to  8/3

t  must be in the interval  \(\Big(-24,\frac83\Big]\)_

See the answer here:  https://web2.0calc.com/questions/help_40221#r1

By the Pythagorean Identity,

\(\sin^2(12^\circ)+\cos^2(12^\circ)\ =\ 1\)

                                                      We are given that   cos(12°) = h   so we can substitute  h  in for  cos(12°)

\(\sin^2(12^\circ)+h^2\ =\ 1\)

                                                      Subtract  h²  from both sides of the equation.

\(\sin^2(12^\circ)\ =\ 1-h^2\)

                                                      Because  12°  is in Quadrant I,  sin(12°)  is positive. So take positive sqrt of both sides.

\(\sin(12^\circ)\ =\ \sqrt{1-h^2}\)

By definition of cotangent,

\(\cot(12^\circ)\ =\ \frac{\cos(12^\circ)}{\sin(12^\circ)}\)

                                          Substitute  h  in for  cos(12°)  and substitute  √[ 1 - h² ]  in for  sin(12°)

\(\cot(12^\circ)\ =\ \frac{h}{\sqrt{1-h^2}}\)_

Thanls!

Let the distance between the center of the base and the northwest corner of the base be  y m

Distance between the center of the base and the west side of the base  =  3/2 m  =  1.5 m

Distance between the center of the base and the north side of the base  =  1.5/2 m  =  0.75 m

By the Pythagorean Theorem,

\((1.5)^2+(0.75)^2\ =\ y^2\\~\\ (\frac32)^2+(\frac34)^2\ =\ y^2\\~\\ \frac94+\frac{9}{16}\ =\ y^2\\~\\ \frac{45}{16}\ =\ y^2\\~\\ y^2\ =\ \frac{45}{16}\)

Again by the Pythagorean Theorem,

\(y^2+2^2\ =\ x^2\\~\\ \frac{45}{16}+4\ =\ x^2\\~\\ \frac{109}{16}\ =\ x^2\\~\\ x\ =\ \sqrt{\frac{109}{16}}\\~\\ x\ =\ \frac{\sqrt{109}}{4}\)___

She has 10 choices   0 - 9     a 1/10 chnace of getting it correct on the first try...    1/10     or 10%

\(cos^2x+sin^2x=1<=> h^2 + sin^212=1<=>sin12= \sqrt{1-h^2}=\sqrt{(1-h)(1+h)}\)

\(cot12=\frac{cos12}{sin12}=\frac{h}{\sqrt{(1-h)(1+h)}}\)

You've asked a zillion of these types of questions now.

Are these for a class?  Are they homework?

What is the reason for posting one after another of these types of problems?

Thank you

Alan is correct as usual.