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Digiron [165]
4 years ago
5

the volume of a prism which has an altitude of 10 units and has a right triangle base with a hypotenuse of 13 units and a leg of

12 units is

Mathematics
1 answer:
AlladinOne [14]4 years ago
8 0
Check the picture below.

since we know the hypotenuse and one of the legs of the triangular base, we can use the pythagorean theorem to get the missing leg, namely the base of the triangle, as you see in the picture.  Thus we end up with a triangle whose base is 5 and height is 12.

now, the volume of the prims, is just the area of its triangular base, times the length, namely 10.

\bf \textit{volume of the triangular prism}\\\\
V=\stackrel{length}{(10)}\stackrel{base's~area}{\left[ \cfrac{1}{2}(5)(12) \right]}

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Can someone help me I have two more other questions they are 15 points each I just posted them right now! Thanks so much
nlexa [21]

Answer:

$10 coins: 21

$20 coins: 24

Step-by-step explanation:

heyy me again

So we can create 2 equations where

x = number of $10 coins

y = number of $20 coins:

x + y = 45

10x + 20y = 690

we can move equation 1 around so that we can get a value of x

x = 45 - y

now we can substitute x into the second equation

10(45 - y) + 20y = 690

450 - 10y + 20y = 690

10y = 240

y = 24

Now plug in y back into the first equation

x = 45 - y

x = 45 - 24

x = 21

8 0
3 years ago
Tell whether the following relation is a function or not<br><br>3<br>. 2
fiasKO [112]
It is a function I think
5 0
4 years ago
A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen
Darina [25.2K]

Answer:

100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving y=x^2 about y axis in the interval 0\leq x\leq 3.

Density of the fluid in the tank, D=100\ lbs/ft^3

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, y(0)=0^2=0

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in x=3 in the parabolic equation . This gives,

H=3^2=9\ ft

So, the height of top of tank is, y(3)=H=9\ ft

Now, 5 ft above 'H' means H+5=9+5=14

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

\Delta y=(14-y) ft

Now, area of cross section of the tank is given as:

A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, y=x^2

So, x=\sqrt y

Therefore, radius, r=\sqrt y

Now, area of cross section is, A(y)=\pi (\sqrt y)^2

Work done in pumping the contents to 5 feet above is given as:

W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy

Plug in all the values. This gives,

W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}

7 0
3 years ago
Solve for x! please helppp!
Colt1911 [192]

Answer:

58°+47°+x=180°[sum of angles in a triangle]

105+x=180

×=108-105

x=75

3 0
3 years ago
What is the improper fraction for 5 1/3
Yakvenalex [24]
Hey there Shawn!

\left[\begin{array}{ccc}\boxed{\boxed{ 5\frac{1}{3}= \frac{16}{3}  }} \ as \ an \ improper \ fraction\end{array}\right]

Hope this helps you budd!
5 0
3 years ago
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