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Andreyy89
3 years ago
15

a right rectangular prism has a base area of 20cm2 and a height of 9.2cm. what is the volume of the prism ??

Mathematics
1 answer:
Tanzania [10]3 years ago
3 0
52.4 because the rectangular prism
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(2x-7)² ?!? I really need some help
sergij07 [2.7K]

Answer:

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(a-b) all square

6 0
2 years ago
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The shapes below are similar. What does x have to be?
laila [671]
I think it is 3.5
if 8 is divided by 4, you get two since they are the same shape, 14 should be divided by 4 as well which is 3.5
7 0
3 years ago
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An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu
exis [7]
Take the derivative with respect to t
- w \sin(wt) + \sqrt{8} w cos(wt)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
0 = -w \sin(wt) + \sqrt{8} w cos(wt)
divide by w
0 =- \sin(wt) + \sqrt{8} cos(wt)
we add sin(wt) to both sides

\sin(wt)= \sqrt{8} cos(wt)
divide both sides by cos(wt)
\frac{sin(wt)}{cos(wt)}= \sqrt{8}   \\  \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\  \\ wt=arctan(2 \sqrt{2)} OR\\ wt=arctan( { \frac{1}{ \sqrt{2} } )
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

I=cos(2n \pi -2arctan( \sqrt{2} ))
since 2npi is just the period of cos
cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 

substituting our second soultion we get
I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))
since 2npi is the period
I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}
so the maximum value =\frac{1}{3}
minimum value =- \frac{1}{3}


4 0
3 years ago
A vet weighs two puppies. The small puppy weighs 4 2/3 pounds. The large puppy weighs 4 2/3 times as much as the small puppy. Ho
Maksim231197 [3]
4 2/3 lbs = small puppy weight
4 2/3 times the small puppies weight = large puppy weight
          4 2/3 × 4 2/3  or 4 2/3^2  = 21.77777.... or in mixed number is 21 7/9 lbs
4 0
4 years ago
Find a quadratic equation whose roots are 1.5 +√2 and 1.5-√2expressing it in the form of ax^2+bx+c+c=0 where a,b and c are integ
Ilia_Sergeevich [38]

0 = f(x) = (x - r)(x - s) = x² - (r+s) + rs

We have r=1.5+√2,  s=1.5 -√2 so r+s = 3 and

rs = (1.5+√2)(1.5 - √2) = 1.5² - (√2)² = 2.25 - 2 = 0.25

f(x) = x² - 3x + -.25

For integer coefficients we mulitply by 4,

g(x) = 4f(x) = 4x² - 12x - 1

Answer:  4x² - 12x - 1 = 0

5 0
3 years ago
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