Solve 5ax + 6 ax = 7ax+ 8 for x. Assume a # 0. O A. x = = B. x = 2a ( C. 2 = D. x = a n

Is (3,2) a solution of the system of equations?Explain 2x+y=8 -x+3y=4

lesya692 [45]

Answer:

no

Step-by-step explanation:

2(3)+2=8

6+2=8

-x+3y=4

-(3)+3(2)=4

-3+6=4

3 does not equal 4

7 0

3 years ago

The number of applicants to a university last year was 11,450. This year, the number of applicants grew by 2%. How

LenKa [72]

Answer:

The number of students on this year = 11679

Step-by-step explanation:

Number of applicants last year = 11450

The growth on this year = 2 %

So the number of students in this year becomes = 11450 × $\frac{102}{100}$

Thus the number of students = 11679

These are the number of students present on this year.

6 0

3 years ago

Solve for n: -8+n=23

BARSIC [14]

Answer: n=31

Reasoning: -8+n=23

First add 8 to both sides (-8+8 and 23+8)

n=31

7 0

3 years ago

Read 2 more answers

Which 2-digit number is equal to the sum of its first digit plus the square of its second digit?

LenaWriter [7]

Let, the two-digit number = xy

It is given that, xy = x + y²

Then, number which obey the rule is 89.

Check: 8 + 9² = 8 + 81 = 89 [ Itself ]

In short, Your Answer would be: 89

Hope this helps!

8 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals:

$0< t$

For this case if we select two values let's say 0.25 and 0.5 we see that

$v(0.25) =6.1875, v(0.5)=3.75$

And we see that for $a=0.5 >0.25=b$ we have that f(b)>f(a) so then the function is decreasing on this case.

$1$

We have a minimum at t=2 since at this value w ehave the vertex of the parabola :

$v_x =-\frac{b}{2a}= -\frac{-12}{2*3}= -2$

And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

v(1.75) =-2.8125< -2.25= v(1.5) so then the function sis decreasing on the interval 1<t<2

$2$

We see that the function would be increasing.

$3$

For this interval we will see that for any two points a,b with $a>b$ we have $f(a)>f(b)$ for example let's say a=3 and b =4

$f(a=3) =0 , f(b=4) =9 , f(b)>f(a)$

The particle is moving to the right then the velocity is positive so then the answer for this case is: $0$

Part c

$a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

5 0

3 years ago

Solve 5ax + 6 ax = 7ax+ 8 for x. Assume a # 0. O A. x = = B. x = 2a ( C. 2 = D. x = a n​

Solve 5ax + 6 ax = 7ax+ 8 for x. Assume a # 0. O A. x = = B. x = 2a ( C. 2 = D. x = a n