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Iteru [2.4K]
4 years ago
11

Can someone help me with this??

Mathematics
1 answer:
timurjin [86]4 years ago
6 0
The volume for a triangular pyramid would be V=1/3AH, where A is the area or the triangular base,  and H would be the height. 
We need to find the area first. We can input the formula of the area of the triangular base, so it would be V=1/3(1/2bh)H. 
In this problem, H=8, b=6, and h=4. Plug that in to the formula. 
V=1/3(1/2*6*4)8. We can put that into a calculator, and get 32 cubic feet as our answer. 
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Which expression represents the volume of the prism?
Tamiku [17]

Answer:

x³-2x²  = x2(x – 2) cubic units

Step-by-step explanation:

The volume of a prism is found by multiplying the base area by the height. The base area is a parallelogram and so the area is x*(x-2) = x² -2x.

Multiply this area by the height x.

V = x(x² - 2x) = x³-2x²

This is the same as x2(x – 2) cubic units.

7 0
3 years ago
Read 2 more answers
The new iPhone 3.14 is being sold at the UT Math Department. The original price is $1200 (day 0), but they are going to decrease
Marat540 [252]

\bf \qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&50.14\\ P=\textit{initial amount}\dotfill &1200\\ r=rate\to r\%\to \frac{r}{100}\dotfill\\ t=\textit{elapsed time}\dotfill &16\\ \end{cases}


\bf 50.14=1200(1-r)^{16}\implies \cfrac{50.14}{1200}=(1-r)^{16}\implies \sqrt[16]{\cfrac{50.14}{1200}}=1-r \\\\\\ r=1-\sqrt[16]{\cfrac{50.14}{1200}}\implies r\approx 0.18\implies \stackrel{\textit{converting to percent}}{r\approx 0.18\cdot 100}\implies r\approx \stackrel{\%}{18}\quad \leftarrow x

3 0
4 years ago
Suppose an airplane traveling in a circle at 138 m/s can turn in a circle with as much as 69.6 m/s of centripetal acceleration.H
Oksanka [162]

Answer: A. 9567m

i hope i'm correct let me know if i'm wrong

6 0
3 years ago
Integrate dx/3sinx+4cosx
german

\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}

A standard approach would be the tangent half-angle substitution:

t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx

Then

\sin x=2\sin\dfrac x2\cos\dfrac x2\implies\sin x=\dfrac{2t}{1+t^2}

\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2\implies\cos x=\dfrac{1-t^2}{1+t^2}

from which we get

\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt

So the integral becomes

\displaystyle\int\frac{\frac2{1+t^2}}{\frac{6t}{1+t^2}+\frac{4(1-t^2)}{1+t^2}}\,\mathrm dt=\int\frac{\mathrm dt}{3t+2(1-t^2)}=-\int\frac{\mathrm dt}{2t^2-3t-2}

Rewrite the denominator as

2t^2-3t-2=(2t+1)(t-2)

and expand the integrand into its partial fractions:

\dfrac1{2t^2-3t-2}=\dfrac15\left(\dfrac1{t-2}-\dfrac2{2t+1}\right)

We have

\displaystyle-\frac15\int\frac1{t-2}-\frac2{2t+1}\,\mathrm dt=-\frac15(\ln|t-2|-\ln|2t+1|)+C

=\dfrac15\ln\left|\dfrac{2t+1}{t-2}\right|+C

=\dfrac15\ln\left|\dfrac{2\tan\frac x2+1}{\tan\frac x2-2}\right|+C

6 0
3 years ago
A particle whose mass is 4 kg moves in xyplane with a constant speed of 2 m/s in the positive x-direction along y = 6 m. Find th
s2008m [1.1K]

Answer: 32.8kgm^2/s

Step-by-step explanation:

1. Angular momentum is a product of linear momentum and the distance of the object from a rotation axis.

2. Linear momentum (L) is given as :

L = mv

Where: m= mass = 4kg

v= velocity= 2m/s

=> L = 8kgm/s

Angular momentum (A) = L x centre distance (d) of object from a rotation axis

d = √(0.9-0)^2 + (10-6)^2 = 4.1m

=> A = 8 x 4.1 = 32.8kgm^2/s.

5 0
3 years ago
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