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NeTakaya
3 years ago
15

The surface area of a triangular pyramid is 1,936 inches squared. If the dimensions are multiplied by 1/4, what will be the new

surface area?
Mathematics
1 answer:
rosijanka [135]3 years ago
8 0

Answer:

121 inches squared

Step-by-step explanation:

If the dimensions are multiplied by 1/4, we have that the surface area will by multiplied by (1/4)^2, as the dimensions are in inches and the surface area is in inches squared.

So, If the original surface area is 1936 inches squared, the new surface area will be:

New surface area = 1936 * (1/4)^2 = 1936 / 16 = 121 inches squared.

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The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 5t 2 i - 4tj - 1k6 lb, and F
olga2289 [7]

Answer:

r=294.9m

Step-by-step explanation:

The forces on the particle are

W=mg\hat{j}\\F_{1}=52\hat{i}+6\hat{j}-2t\hat{k}\\F_{2}=5t^{2}\hat{i}-4t\hat{j}-1\hat{k}\\F_{3}=(5-2t)\hat{i}

Now , we sum all these forces to get the net force

F_{T}=W+F_{1}+F_{2}+F_{3}\\F_{T}=(52+5t^{2}+5-2t)\hat{i}+((6+6-4t)\hat{j}+(-2t-1)\hat{k}\\F_{T}=(57-2t+5t^{2})\hat{i}+(12-4t)\hat{j}+(-2t-1)\hat{k}\\

we can use the fact F=m*a and integrate the acceleration

a(t)=\frac{1}{m}F(t)\\\\v(t)=\int a(t)dt=\frac{1}{m}\int{F_{T}}dt\\\\v(t)=\frac{1}{m}[(57t-t^{2}+\frac{5}{3}t^{3})\hat{i}+(12t-2t^{2})\hat{j}+(-t^{2}-t)\hat{k}]\\\\r(t)=\int v(t)dt=\frac{1}{m}[(\frac{57}{2}t^{2}-\frac{1}{3}t^{3}}+\frac{5}{4}t^{4})\hat{i}+(6t^{2}-\frac{2}{3}t^{3})\hat{j}+(-\frac{1}{3}t^{3}-\frac{1}{2}t^{2})]

and we evaluate in r(2) an we take the norm to obtain the distance

r(2)=\frac{1}{m}[\frac{394}{3}\hat{i}+\frac{56}{3}\hat{j}-\frac{14}{3}\hat{k}]\\|r(2)|=\frac{1}{m}\sqrt{[(\frac{394}{3})^{2}+(\frac{56}{3})^{2}+(\frac{14}{3})^{2}]}\\|r(2)|=\frac{132.73}{0.45}=294.9m

I hope this is useful for you

regards

8 0
4 years ago
If 6 men build a wall in 15 days, then how many days will 9 men take?
Lisa [10]
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5 0
3 years ago
Find the domain over which the function y = x + 6x is monotonic increasing.
iragen [17]

Answer:

x > -3

\sf (-3, \infty)

Step-by-step explanation:

Domain: input values (x-values)

Monotonic increasing:  always increasing.  
A function is increasing when its graph rises from left to right.

The graph of a quadratic function is a parabola.  If the leading term is positive, the parabola opens upwards.  The domain over which the function is increasing for a parabola that opens upwards is values greater than the x-value of the vertex.

<u>Vertex</u>

Standard form of quadratic equation:  \sf y=ax^2+bx+c

\textsf{x-value of vertex}=\sf -\dfrac{b}{2a}

Given function:

\sf y=x^2+6x

Therefore, x-value of function's vertex:

\sf \implies x= -\dfrac{6}{2}=-3

<u>Final Solution</u>

The function is increasing when x > -3

\sf (-3, \infty)

4 0
3 years ago
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