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10^x=76333492

xlog(10)=log(76333492)

x = log(76333492) = 7.882
so 8

a)
initial velocity in the y direction = v₀sinθ = 70sin(30°)=35m/s
bc of symmetry and parabola stuff, final velocity in the y direction = -35m/s

use v=v0+at, solve for t ---> t = (v-v0)/a = (-35m/s-35m/s)/(-9.8m/s²) = 7.14286s = 7.14s

b)

initial velocity in the x direction = v₀cosθ = 70cos(30°)=60.62178m/s

you found the time it takes to hit the ground in part a, so use that in this equation:
d = vt, where d = distance/displacement, 
d = 60.62178m/s * 7.14286s = 433.01ft

Thank you! I have a question though, How did you get that answer? It seems like we don't have enough information?

ok, here is my logic,

No promises that it is correct.

Give one property to each child.  9*8*7 = 504 ways.

Now there are 6 properties left to be divided between the 3 children

6,0,0  = 3 ways

5,1,0 = 3!*6 = 36ways

4,1,1 = 3!*6*5/2 = 90ways

4,2,0 =  3!*6C4 = 90ways

3,3,0 = 3*6C3 = 60ways

3,2,1 = 3!*6C3*3 = 360ways

2,2,2 = 6C2*4C2 = 90ways

3+36+90+90+60+360+90 = 729ways

9*8*7*729 = 367,416 ways

I am certainly not going to guarantee that this is correct.

If you get the correct answer guest, then please share it with us. Preferably with the working.

Well, we just make h(t) 0 since it reprsents the height.

-16t^2 + 100t + 5 = 0

Using the Quadratic Formula, 

We get the value of t as 25/8 +/- sqrt(645)/8.

The two solutions approximated are -0.05 seconds and 6.23 seconds. We take the positive solution in this case. It takes 6.23 seconds for the ball to reach the ground.

Let's first get this into a normal equation "y = mx + b" form.

-3x + 3y >= 12

3y >= 3x + 12 

y >= x + 4

From this, we can find the x and y intercepts to make it easier to graph. After plugging in 0 for x and y, we get the two intercepts....

x - intercept: (0, -4)

y - intercept: (4, 0)

Then, since it is >= and not just >, the line is not dotted, but it is instead shaded in because those are included in the set of values for which y >= x + 4 is valid. Also, since it is >=, you shade above the line. From all of these, I can conclude it is Graph #4.

Total possibilities with no restrictions

12 steps altogether. 5 are horizontal   12C5=792  just as you got

Total number that go to (4,3)

7 steps to get from start to (4,3) four are horizontal = 7C4 = 35 ways

Number of ways from (4,3) to (5,7)

5 steps, 1 horizontal = 5 ways

35*5=175 ways to get from start to finish if you go through (4,3)

Number of ways to get from start to finish if you do not go through (4,3) is    792-175 = 617

P(x)  =  a(x + b)²(x - c)

point (-1, 0) is a zero with multiplicity 2   --->   b = 1

point (4, 0) is on the graph   --->  c = 4

P(x)  =  a(x + 1)²(x - 4)

contains the point (0, -12)   --->   -12  =  a(0 + 1)²(0 - 4)

                                                     -12  =  a(1)(-4)

                                                        3  =  a

https://web2.0calc.com/questions/helpppp_30

Area of a Triangle  =  ½ · base· height

Using 15 for the base and 8 for the height (because they are perpendicular), 

the area is:  ½ · 8· 15  =  60

Now, using 17 for the base and x for the height, and we know that the area of the triangle is 60:

     60  =  ½ · 17· x

     60  =  8.5 · x

     7.06  =  x  (approximately)  (using common fractions, the height is  120/17)

arc(TA)  =  2·angle(QTA)   --->   arc(TA)  =  2·41  =  82

arc(AU)  =  2·angle(RUA)   --->   arc(AU)  =  2·63  =  126

arc(TAU)  =  arc(TA) + arc(AU =  82 + 126  =  208

arc(TU)  =  360 - arcTAU)  =  360 - 208  =  152

angle(QPR)  =  ½( arc(TAU) - arc(TU) )  =  ½( 208 - 152 )  =  ½(56)  =  28

a)
\(Area = \int_a^bf(x)dx\\ \int_1^3\sqrt{x-1}dx=\frac{4\sqrt{2}}{3}\)

b)

\(Average\;Value = \frac{1}{b-a}\int_a^bf(x)dx\\ \frac{1}{3-1}\int_1^3\sqrt{x-1}dx=\frac{1}{2}*\frac{4\sqrt{2}}{3}=\frac{2\sqrt{2}}{3}\)

c)

\(Volume = \int_a^bA(x)dx=\int_a^b\pi r(x)^2dx=\pi\int_a^br(x)^2dx\\ \pi\int_1^3(\sqrt{x-1})^2dx=2\pi\)

Hello Guest!

Let x and y be the two integers. 

x/y = 21 ---- 1

Let x>y

x-quotient = 21

Quotient =21,

so x-21=21 --- Add 21 to both sides

x=42

By Equation 1 --> 42/y=21 --- Multiply by y on both sides

42=21y, divide by 21 on both sides.

y=2 

So, the sum is x+y = 42+2 = 44

Hope this helps Guest!

What do you mean?

The minimum possible value of \(\Box\) is 173.

The hypotenuse is 4*sqrt(26).

The Answer is B. 

Since at the point of intersection, the two equations will have the same values of x and y, we set the two equations equal to each other. This gives an equation that we can solve for x

Then we substitute that x value in one of the line equations (it doesn't matter which) and solve it for y.

(That is how you solve for the intersection.)

Hope this helps you learn!

\(x\) is the number of hours Lenny worked in July and \(\frac{1}{2}x\) is the number of hours that Lenny spent working in June.

We know that Lenny worked 84 hours, so we add the two coefficients to get \(1.5x = 84 \)

This means that in the month of July, Lenny worked 56 hours, and in the month of June, Lenny worked 26 hours. 

\(x=56 \) and \(1.5x = 28\)

Hope this helps!

Consider ΔABC with AB = 5 and AC = 6. Let D be on point AC such that CD = 2AC  and ∠CBD =2 ∠ABD.

Find length AC.

Text in red and purple color needs correction.   

It's solved below ...

length  =  L

width  =  W

height  =  H

length = twice the width        --->   L  =  2W   

length  = 3 less than height   --->   L  =  H - 3     

Since  L  =  2W  and  L  =  H - 3   (combining them)   --->   H - 3  =  2W   --->   W  =  (H - 3) / 2

The total surface area is 100   --->   2LW + 2LH + 2 HW  =  100   --->   LW + LH + HW  =  50

Substituting:   ( H - 3 )( (H - 3)/2) + ( H - 3 )(H) + (H)( (H - 3)/2 )  =  50   

--->    ( H² - 6H + 9 ) / 2  +  H² - 3H  + ( H² - 3H ) / 2  =  50

Multiplying by 2:

--->   H² - 6H + 9 + 2H² - 6H + H² - 3H  =  100

--->   4H² - 15H - 91  =  0

--->   (H - 7)(4H + 13)  =  0

--->   H  =  7

1/a  + 1/b  + 1/c = 1/2

1/6  +  1/4 + 1/c  = 1/2       solve for 'c' the time of chad alone

2/12  + 3/12  + 1/c = 1/2

1/c  =  1/2 - 5/12

1/c = 1/12

c = 12 hours

Common ratio is 3r

For the geometric series to converge, the absolute value of the common ratio has to be less than 1

|Common ratio| < 1
|3r| < 1
-1/3 < r < 1/3
or

(-1/3,1/3)

I entered that in and it didn't work,  Can you check again?

c = child tix                     revenue from = $ 6 c

a = adult tix = 166-c       revenue from = $ 9.70(166-c)   Summed, the two revenues total 1351,20

6 c   + 9.7 (166-c )    = 1351.20      solve for 'c' the number of chlid tix