Can someone plz help plz show work

Mathematics

1 answer:

aliya0001 [1]3 years ago

8 0

Answer:

Step-by-step explanation:

In the base we have the right triangle. Use the Pythagorean theorem for to calculate the hypotenuse:

$h^2=4^2+3^2\\\\h^2=16+9\\\\h^=25\to h=\sqrt{25}\\\\h=5\ ft$

The Lateral Area is the sum of the areas of three rectangles:

$4\ ft\ \times\ 2\ ft\to A_1=(4)(2)=8\ ft^2\\\\3\ ft\ \times\ 2\ ft\to A_2=(3)(2)=6\ ft^2\\\\5\ ft\ \times\ 2\ \ft\to A_3=(5)(2)=10\ ft^2$

The Lateral Area:

$LA=A_1+A_2+A_3\to LA=8+6+10=24\ ft^2$

You might be interested in

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

What is the surface area of a cylinder with a radius of 3ft and a height of 5ft

ycow [4]

The answer is 150.8. there is a cylindrical surface area calculator online if you have anymore related questions

4 0

3 years ago

Read 2 more answers

What is the answer to this question

timama [110]

Answer:

$b^6z^612a^5$

Step-by-step explanation:

The first step is to combine like terms, and multiply them together first. Since multiplication is commutative, it doesn't matter in what order you do it. Therefore, this can be rewritten as $(2a^2\cdot 6a^3)\cdot(b^4\cdot b^2)\cdot(z\cdot z^5)=(12a^5)\cdot(b^6)\cdot(z^6)=b^6z^612a^5$ . Hope this helps!