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\(\frac{2}{3} t - 1 < t + 7 \le -2t + 15 \)

We first have  the inequality

2/3 t - 1 < t + 7    subtract 7 from both sides

2/3 t  -  8 <  t     subtract 2/3 t  from both sides

-8 < 1/3 t     multiply both sides by 3

-24 <  t

And we have the second inequalty

t + 7 ≤  -2t + 15       add 2t  to both sides...subtract 7 from both sides

3t ≤  8      divide both sides by 3

t ≤ 8/3

So      (-24, 8/3 ]    is the solution

See the pic, here :

Since ABC = 49  and ACB  = 12

Then BAC  =  180  - 49 - 12 =  119

Then  we have quadrilateral HFAE  

Angle HFA  = 90

Angle FAE  =  Angle BAC = 119

Angle AEH  = 90

So  angle FHE  = Angle BHC  =   360 - 90  - 90 - 119   =  61°

Let the weight of a sack of sugar  =  x

Let the weight of a sack of flour  = y

We have these inequalities

3x + 4y ≤  50

2y - 3x ≤ 13

And we wish to maximize y

Look at the graph, here :

https://www.desmos.com/calculator/aiwomrcer1

The  value that maximizes y, the weight of a sack of flour,  is   10.5 lbs

OK

So far  your team has played  20 games.......

And the number you have won =  3

So.....let x  be the number of additional weeks....since you win 3 games a week...you will win 3x additional games....and you will play 4x additional games

[ 3 + 3x]

_______   =  .65     cross-multiply

[ 20 + 4x]

3 + 3x  =  .65 [ 20 + 4x]      simplify

3 + 3x  =  13 + 2.6x      subtract 3, 2.6x  from both sides

.4x  =  10      divide both sides by .4

x  =  25   additional weeks..so...since you have already played 5 weeks...you need to play

25  +  5   =  

30 total weeks to get to a winning percentage of 65%

Proof

[3 + 3(25)]                78

_________   =      ______   =  .65

 20 + 4(25)               120

A) You can use the same diagram.....just add segment HE

This segment will form one leg of  right triangle HEJ   = 3960 miles

And JE forms the hypotenuse  = GE + JG =   3960 + 6  =  3966 miles

B)   So....using the Pythagorean Theorem

JH  = 

sqrt ( JE^2 - HE^2 )  =

sqrt  (3966^2 - 3960^2 )   =

sqrt (47556)  ≈

C)   218.1 miles = distance to the horizon 

We can use some symmetry to solve this

Draw a segment from the center of the square to the top right vertex of the square...this will divide the shaded area into to 2 equal parts...let this segment  be EC

If we let the center of the the circle on the right be the midpoint of BC  ....call this point, F....and the radius of this circle will  = 4

Then.....  the sector FEC of the circle on the right   will be the quarter area of this circle  = 

(1/4)pi (4)^2  =  4pi  units^2

And.....  the triangle formed by FEC  wll have the area  (1/2)(4)^2sin 90  = 8 units^2

So the top half of the shaded area is just the difference of these two areas  =

[ 4pi  -  8]    units^2

So....the whole shaded area is just twice this  =

[ 8pi - 16 ] cm^2  ≈   9.13 cm^2

Here's a pic, tertre :

Thanks so much, hectictar!

Thank you so much, hectictar!

A  = 30.5°

c  =  52.8

Angle C  =  90 - 30. 5   =  59.5°

Using the Law of Sines

a / sin A  =  c / sin C

a / sin 30.5 =    52.8 / sin 59.5          multiply both sides by sin 30.5

a =  52.8*sin 30.5  / sin 59.5    =  31.1 ft

OB   =   OC + CB   =   32 + 36   =   68

AB   =   2 * OB   =   2 * 68   =   136

AC   =   AB - CB   =   136 - 36   =   100

diameter of circle K   =  AC  =  100

radius of circle K   =  100/2   =   50

area of semicircle K   =   (1/2)(pi)(50²)   =   1250pi  sq units

If I can read this...

angle B  =  62°

b  = 26.5

So A  =  90  - 62  =  28°

So......using the Law of Sines , we have

a/ sin A  =  b / sin B

a / sin 28 = 26.5 / sin 62        multiply both sides by sin 62

a  =  26.5 * sin 28  / sin 62  =   14.1 ft

Thanks so much, CPhill! Wonderful!

We know angle B is 38°.

Side  a  is adjacent to angle B.

We know the hypotenuse is 15 ft.

And....

cos( angle )  =  adjacent / hypotenuse        And in this case...

cos 38°   =   a / 15

So the cosine function should be used.

Thanks

The side opposite angle B  is  AC   =   "z"

The area, A, of a trapezoid is given by :

A =  height *  ( sum of the base lengths) / 2

So we have

123.265 = 8.9 * ( 17.8  + missing base)  / 2

Multiply both sides by 2

247.25  =  8.9 * (17. 8 + missing base ) 

Divide both sides by 8.9

27.78   =  17.8 + missing base   

Subtract 17.8  from both sides

≈ 9.98 miles  = missing base

The interest  is

Final value  - Amount invested  =

15000 (1.2)(6/3)  - 15000 =

15000 (1.2)(2) - 15000 =

15000 (2.4 - 1)  =

15000 (1.4)  =

$21000

I think you made a typo in your math: 140+25+60 = 225 students who participate in all

330 - 225 = 105 - students who do not participate.

105 / 330 = 7 / 22 - probability that a randomly picked student is a non-participant.

If both of them are accurate, why is there a difference of: 22.274 - 21.991 =0.283 m^2 ???!!!.

translation to english from portuguese:

suppose that the equation: y 985.3 + 0.74 X represents the adjusted straight line of a metal bar and its length, based on in this equation, estimate the length value when the temperature equals 41 ° C.

Let's look at AZSun's answer a litlle more in detail.....

If Betty  and Wilma  join the group it is still not divisible into equal size groups

Note that all the even integers  from 30 - 50  can always be divided into equal groups because, at least, they are all divisible by 2

But look at the odd integers between  30 - 50

31   33    35    37   39    41   43   45   47  49

We are looking for two adjacent prime numbers

These  are   41  and 43

Note....41 students cannot be divided into equal groups and even if  Betty and Wilma join in....the group size  = 43    which isn't capable of being divided equally, either  !!!!

So...AZSun's answer of 41 is good  !!! 

Thank You!!!, Cphill

Chuck can play in the three quarters of the circle with radius 3 that does not have any area inside the shed.

The area of that is \(\frac {3}{4} \times9 \pi = \frac{27\pi} {4}\).

One meter of Chuck's rope can extend below the shed on the bottom right corner with one meter left. So, Chuck has a quarter circle with radius \(1\) he can play in.

The total is \(\frac{27\pi} {4} + \frac{\pi}{4}= \frac{28\pi}{4} = \boxed{7\pi}\) .

area of rectangle=length*width

area of rectangle=3*2

area of rectangle=6m^2

area of circle=π*r^2

area of circle=π*3^2

area of circle=π*9

area of circle=28.27433388m^2

area of circle - area of rectangle= area around the outside of the shed

28.27433388m^2-6m^2=22.27433388m^2

(a): Median is the middle score, when arranged in order. So, 140