Write an algebraic expression that is equivalent to the expression.<br><br> cos⁡(arcsin⁡ (x-h)/r )

A metal conduit will be used as a pathway for wiring through a concrete block. The conduit is a 5 foot long rod with an outer di

kolbaska11 [484]

Answer: 586.59 cubic centimeters .

Step-by-step explanation:

As per given . we have

Inner diameter = 1.8 inches

⇒Inner radius :r = 0.9 in. (radius is half of diameter)

= 0.9 x (2.54) = 2.286 cm [∵ 1 in . = 2.54 cm]

Outer diameter = 2 inches

⇒Outer radius : R = 1 inch = 2.54 cm

Height : h = 5 feet = 5 x(30.48) = 152.4 cm [∵ 1 foot = 30.48 cm]

The formula to find the volume of a hollow cylinder :

$V=\pi(R^2-r^2)h$ , where R= outer radius , r= inner radius and h= height.

Now , the volume of metal in the conduit :

$V=(3.14)(( 2.54)^2-(2.286)^2)(152.4)$

$V=(3.14)(6.4516-5.225796)(152.4)$

$V=(3.14)(1.225804)(152.4)$

$V=586.591342944\approx586.59\ cm^3$

Hence, the volume of metal in the conduit is 586.59 cubic centimeters .

5 0

3 years ago

THERE ARE 3 BAGS OF APPLES WEIGHING A TOTAL OF 22 1/2 POUNDS. TWO OF THE BAGS WEIGH 6 3/8 POUNDS AND 3 1/4 POUNDS. HOW MUCH DOES

Mkey [24]

I wasn't going to click on this one, but the all-caps enthralled me and hypnotized me.

first off, let's change all the mixed fractions to "improper" and proceed from there, keep in mind that if we subtract the two bags' weight from the total, what's leftover is the 3rd bag's weight.

$\bf \stackrel{mixed}{22\frac{1}{2}}\implies \cfrac{22\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{45}{2}}
\\\\\\
\stackrel{mixed}{6\frac{3}{8}}\implies \cfrac{6\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{51}{8}}
\\\\\\
\stackrel{mixed}{3\frac{1}{4}}\implies \cfrac{3\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{13}{4}}\\\\
-------------------------------\\\\$

$\bf \stackrel{\textit{sum of the two bags}}{\cfrac{51}{8}+\cfrac{13}{4}}\impliedby \textit{our LCD is 8}\implies \cfrac{(1)51+(2)13}{8}
\\\\\\
\cfrac{51+26}{8}\implies \cfrac{77}{8}\\\\
-------------------------------\\\\
\stackrel{total}{\cfrac{45}{2}}~-~\stackrel{two~bags}{\cfrac{77}{8}}\impliedby \textit{our LCD is again 8}\implies \cfrac{(4)45-(1)77}{8}
\\\\\\
\cfrac{180~~-~~77}{8}\implies \cfrac{103}{8}\implies \stackrel{third~bag}{12\frac{7}{8}}$

3 0

3 years ago

The scale on a map is 1 in. :25 mi. Your measure 6.5 in. Between two towns. What is the actual distance?

pshichka [43]

Answer:

162.5mi

Step-by-step explanation:

If the scale of the map is 1in:25mi, and you measure 6.5in between two twoms, then the real distance would be 6.5in*25mi = 162.5mi.

6 0

3 years ago

PLEASE HELP ME ASAP!!!! A circle fits in a square with sides of 12 inches. What is the approximate area of the shaded region bet

den301095 [7]

Answer:

Depending on the shaded area, it will be less than 113 in², so not the last option

Step-by-step explanation:

Area of a circle is A= πr²

if it fits in 12 inch sides, 12 i the diameter and 6 is the radius

A= 3.14(6)²

A= 113 in²

Then, find what the shaded area is

4 0

3 years ago

List 3 values that would make this inequality true. 65<15x

Romashka-Z-Leto [24]

Anything 5 and up, because when you multiply the by 15, they equal more than 65, making the inequality true.

5 0

3 years ago

Write an algebraic expression that is equivalent to the expression. cos⁡(arcsin⁡ (x-h)/r )