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Montano1993 [528]
4 years ago
5

Can someone answer #13, #15, and #16

Mathematics
2 answers:
Dominik [7]4 years ago
8 0
#13: b
#14: c
#16: c

It is very easy, have a nice day :)
Mars2501 [29]4 years ago
3 0
16). Volume of a cube: v = s³         (s) stands for side length

v = 3³
v = 3 × 3
v = 9

You were right, the answer is a.
__________________________________________________________
 
I drew a picture to figure out this problem. I drew a circle with 16 different sections. Half of them were black and the other half were white. If a dart is thrown 3 times and lands on the black each time, then you have to eliminate the 3 times. So on the picture, eliminate 3 black blocks. The question wants to know the probability of the dart landing on a black if it's thrown a fourth time. In order to find this out, you need to eliminate just one more black section. Then, you have four more black sections left. You have to put 4 over the number of total black sections there were, which are 8. 4/8 = 1/2, so the final answer is 50%. Sorry if this is confusing. 

__________________________________________________________


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Max is determined the volume of the triangle pyramid below.
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Which fraction is closer to 1/2 than 1?<br> 7/8<br> 10/12<br> 12/14<br> 7/10
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7/10

Step-by-step explanation:

7/8 = .875

10/12 = .833

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What is the result when 4x2 + 7x − 5 is subtracted from 9x2 − 2x + 3?
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Find the length of the following​ two-dimensional curve. r (t ) = (1/2 t^2, 1/3(2t+1)^3/2) for 0 &lt; t &lt; 16
andrezito [222]

Answer:

r = 144 units

Step-by-step explanation:

The given curve corresponds to a parametric function in which the Cartesian coordinates are written in terms of a parameter "t". In that sense, any change in x can also change in y owing to this direct relationship with "t". To find the length of the curve is useful the following expression;

r(t)=\int\limits^a_b ({r`)^2 \, dt =\int\limits^b_a \sqrt{((\frac{dx}{dt} )^2 +\frac{dy}{dt} )^2)}     dt

In agreement with the given data from the exercise, the length of the curve is found in between two points, namely 0 < t < 16. In that case a=0 and b=16. The concept of the integral involves the sum of different areas at between the interval points, although this technique is powerful, it would be more convenient to use the integral notation written above.

Substituting the terms of the equation and the derivative of r´, as follows,

r(t)= \int\limits^b_a \sqrt{((\frac{d((1/2)t^2)}{dt} )^2 +\frac{d((1/3)(2t+1)^{3/2})}{dt} )^2)}     dt

Doing the operations inside of the brackets the derivatives are:

1 ) (\frac{d((1/2)t^2)}{dt} )^2= t^2

2) \frac{(d(1/3)(2t+1)^{3/2})}{dt} )^2=2t+1

Entering these values of the integral is

r(t)= \int\limits^{16}_{0}  \sqrt{t^2 +2t+1}     dt

It is possible to factorize the quadratic function and the integral can reduced as,

r(t)= \int\limits^{16}_{0} (t+1)  dt= \frac{t^2}{2} + t

Thus, evaluate from 0 to 16

\frac{16^2}{2} + 16

The value is r= 144 units

5 0
3 years ago
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