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From the information given above, we can say that the width of the rectangle is \(4\)

The area of a rectangle is \(length \cdot width\)

That is, \(4 \cdot length = 16\)

That gives us length = \(16/4 = 4\)

From that, we can tell that the value of \(x\) must be \(4\).

I think you might have assumed a right triangle with legs of 8 and 12....in that case, your answer would be correct 

Sorry!

Its more complicated than I thought!!!

First, its good to know that triangle PRQ and MNP are similar 

Angle PQR=RPQ are 58deg

Since PRQ and MNP are similar, both angles RPQ and NPM are 58deg

Since this is an isosceles triangle...

58deg+2(anglePMN)=180

So we get...

2(anglePMN) = 122

Angle PMN = 61 degrees

The semi-perimeter of this  triangle =  (8 + 12 + 17) /2  =  37/2  = 18.5

The area of this triangle   = sqrt  ( 18.5 * ( 18.5- 8) (18.5  -12) ( 18.5 - 17) )  =

sqrt [ 18.5 * 10.5 * 6.5 * 1.5 ]  = 

sqrt [  1893. 93.75 ]

The longest two altitudes will be drawn to the shortest two sides

So, using the area formula for a triangle,   we  have

sqrt (1893.9375 )  = (1/2)(8)  *altitude  1

sqrt(1893.9375) / 4    =    altitude 1      (1)

And

sqrt (1893.9375 ) = (1/2)(12)  * altitude 2

sqrt (1893.9375 ) / 6   = altitude 2          (2)

Adding  (1) and (2)  we get

sqrt (1893.9375) / 4  +  sqrt(1893.9375) / 6    ≈  18.133 units

Mrs. Read uses \(3^3 \pi {4 \over 3} = 36 \pi\) inches of yarn per mitten. 

The bigger ball has a volume of \(9^3 \pi {4 \over 3} = 972 \pi\) inches of yarn per mitten. 

Thus, she can make a total of \({972 \pi \over 36 \pi} = \color{brown}\boxed{27}\) mittens. 

\(8+12=20\)

So the sum of the lengths of the two longest altitudes in that triangle is \(20\)

Kinda sus man!!!!

Excellent  !!!!

Let's go through this

Slant height  =18

The side of the base =  sqrt (288)  =  12sqrt 2

The diagonal  length of the base  = 12sqrt (2)  * sqrt (2)  =  24

Half of this diagonal length  =  12

Using the Pythagorean Theorem, the  height of the pyramid  =  sqrt  [  slant height ^2  - (half of diagonal length)^2 ]   = 

sqrt [ 18^2 - 12^2 ] =  sqrt (180)  =6 sqrt (5)

The  volume of the pyramid  =   (1/3) ( base area) (height)  = 

(1/3) (288) (6 sqrt (5) )  =

2 * 288 * sqrt (5)  =

576sqrt (5) cm^3  ≈  1287.975 cm^3    (just as qjin27 found  !!!! )

Answer: if 1 + 1 = 2 then a + b = c therefore I don't know what your answer is buuuuuuuuuut If you use this then you COULD find the answer https://tinyurl.com/58eadpaj  

Hopefully, this will help:

https://web2.0calc.com/questions/counting_31398

The volume of a cone is 1/3hπr²

The height is 15 and the radius must be 7.5.

Substituting these values in the formula gives 

1/3*15*π*(7.5)^2

This gives approx. 883.57

‎‎‎‎‎‎‎‎‎

Very nice, Vinculum  !!!!!

1.  The limit as we approach -4 from the left =   +inf  = undefined

2.  The limit as we approach -4 from the right  =  -inf  =  undefined

3. The limit as we approach -2 from both sides  = + inf  =undefined

4. The limit as x approaches + inf  =  -1      (this is the horizontal asymptote )

5. The limit as x approaches - inf =  undefined    (the function oscillates in this direction )

\(\left|x-5\right|+11\le \:18\)

\(\left|x-5\right|+11-11\le \:18-11\)

\(\left|x-5\right|\le \:7\)

\(\mathrm{Apply\:absolute\:rule}:\quad \mathrm{If}\:|u|\:\le \:a,\:a\:>\:0\:\mathrm{then}\:-a\:\le \:u\:\le \:a\)

\(-7\le \:x-5\le \:7\)

\(x\ge \:-2\quad \mathrm{and}\quad \:x\le \:12\)

\(== -2\le \:x\le \:12\)

Area of cylinder: \(2\pi r^2+2\pi rh\)

\(2\pi \:\:r^2+2\pi \:\:r\left(2.5r\right)\)

\(7\pi \:r^2=448\pi \)

\(r=-8,\:r=8\)

But a radius can't be negative so our radius is \(8\)!!!

This might not be the easiest approach....but.....

By Leg - Leg ,  the two right triangles   are congruent.....so DC = DB   =11

Since angle ABC is bisected, then AB  =(25/11)BC 

And, since angles ADB and CDB are supplementary  the  cosine of  ADC  = - cosCDB

By the Law of Cosines

(  BC )^2  = 11^2  + 11^2  - 2 (11)(11) (cos CDB)

(BC)^2  = 242   - 242 ( cos CBD)

[(BC)^2 - 242 ] / [ -242]   = ( cos  CBD )         (1)

And

[ (25/11)BC]^2  = 11^2 + 25^2  - 2 (11)(25) (-cos CBD)

[ (625 /121)BC^2 ] = 746 + 550 (cos CBD)

[ (625/121)BC^2  - 746 ] / [550]  =  (cos CBD)   (2)

Equate  ( 1) and (2)

[ (BC)^2 - 242 ] / [ -242]  =  [ (625 /121) BC^2 - 746 ] / 550         cross-multiply

550 [ (BC)^2  - 242 ] =  (-242)[ (625/ 121) (BC^2) - 746]        simplify

550BC^2  - 133100 =  -1250BC^2 + 180532

1800BC^2  = 313632

BC^2  = 313632 / 1800

BC  =  sqrt [ 313632 / 1800 ]  =  13.2

So   AB  = (25/11) (13.2)   =  30

The semi-perimeter of   ABD  =  (25 + 11 + 30) / 2  =  33

So  ....using Heron's Formula, [ ABD ]  is

sqrt  [ 33 * ( 33-25) * (33 - 11) * ( 33-30) ] = 

sqrt [ 33 * 8 * 22 * 3]  =

sqrt [ 17424 ]   =

132  units ^2

\(g(x) = \sqrt{16+9}=5\)

Sub \(5\) into function \(t(x)\):

\(3-\sqrt{5+9}\) \(=3-\sqrt{14}\)

Answer is \(3-\sqrt{14}\)