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3 years ago

How do I do this? It is an algebraic expression I believe and

Mathematics

1 answer:

FinnZ [79.3K]3 years ago

6 0

The Answer would be A. First you have to distribute the 5 to the 2x and the 11 and you get 10x+55. then you would combine the 10x and the -3x and get 7x then you would combine the 55 and the 5 by adding and get 60 so then your final asnwer would be A. 7x+60

You might be interested in

Which of the following represents 8 square root x5

gavmur [86]

Answer:

$5 \sqrt{8}$

6 0

3 years ago

A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,

zloy xaker [14]

Answer:

$t=-20ln\left(\dfrac{1}{3}\right)$

Step-by-step explanation:

The relationship between A, the area of the glacier in square kilometers, and t, the number of years the glacier has been melting, is modeled by the equation.:

$A=45e^{-0.05t}$

We want to determine the value of t for which the area, A(t)=15 square kilometers.

$15=45e^{-0.05t}\\$Divide both sides by 45\\\dfrac{15}{45} =\dfrac{45e^{-0.05t}}{45}\\\dfrac{1}{3}=e^{-0.05t}\\$Take the natural logarithm of both sides\\ln\left(\dfrac{1}{3}\right)=ln\left(e^{-0.05t}\right)\\ln\left(\dfrac{1}{3}\right)=-0.05t\\$Divide both sides by -0.05$\\t=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05} \\=-\dfrac{ln\left(\dfrac{1}{3}\right)}{0.05}\\t=-20ln\left(\dfrac{1}{3}\right)$

Therefore, the time for which the area will be 15 sqyare kilometers is:

-20 ln(1/3) years.

7 0

3 years ago

If two objects travel through space along two different curves, it’s often important to know whether they will collide. (Will a

galina1969 [7]

Answer:

Both objects collide at t=3 in the point <9,9,9>

Step-by-step explanation:

Collision Of Moving Objects

Two objects can describe different trajectories in the space. Those trajectories can intersect in one or more points but it doesn't mean they collide. Collision occurs if they are in the same position at the same time. If we know the positions as a function of time of each object, we could try so find if, for a given time, they are in the same position.

The positions of two object are given as

$r1(t)=$

$r2(t)=$

Let's find out if there is at least one value of t that makes both positions to be the same. We can try by equating one of the three coordinates and testing if the value of t make both have the same x,y,z coordinate. Let's try equating the x-components of both

$t^2=4t-3$

Rearranging

$t^2-4t+3=0$

Factoring

$(t-1)(t-3)=0$

We found two solutions

$t=1,\ t=3$

for t=1 the x-coordinates are

$x1=t^2=1$

$x2=4t-3=1$

For t=3

$x1=t^2=9$

$x2=4t-3=9$

Now we'll test both values in the y-coordinates

$y1=7t-12$

$y2=t^2$

For t=1

$y1=-5$

$y2=1$

Thus they don't collide at t=1. Let's try t=3

$y1=7(3)-12=9$

$y2=3^2=9$

Now let's try the z-coordinate for t=3

$z1=t^2=9$

$z2=5t-6=9$

Since the three coordinates match, we can say both objects collide at t=3 in the point <9,9,9>

6 0

4 years ago

1. Solve for xº. 52° can someone help me please

OLEGan [10]

Answer:

x=38°

Step-by-step explanation:

x° and 52 make up a right angle, which is 90°, so for an equation you can do

x+52=90

subtract 52 on both sides and you get

x=38

5 0

3 years ago

Read 2 more answers

uppose that we have a function with a constant amount of work done in initialization, a call to a log-linearsorting algorithm, a

White raven [17]

Answer:

The running time is quadratic (O(n²) )

Step-by-step explanation:

For the set up, we have a constant running time of C. The, a log-linearsorting is called, thus, its execution time, denoted by T(n), is O(n*log(n)). Then, we call n times a linear iteration, with a running time of an+b, for certain constants a and b, thus, the running time of the algorithm is

C + T(n) + n*(a*n+b) = an²+bn + T + C

Since T(n) is O(n*log(n)) and n² is asymptotically bigger than n*log(n), then the running time of the algorith is quadratic, therefore, it is O(n²).

7 0

3 years ago