All categories

3 years ago

If $x$, $y$, and $z$ are positive integers such that $6xyz+30xy+21xz+2yz+105x+10y+7z=812$, find $x+y+z$.

Mathematics

1 answer:

harina [27]3 years ago

8 0

Answer:

Step-by-step explanation:

When we simplify we get $z(6xy+21x+2y+7)+30xy+105x+10y=812.$

Then we continue to factor to get: $(z+5)(6xy+21x+2y+7)&=847.$

We then see that we can factor $(6xy+21x+2y+7)$ into $(z+5)(3x+1)(2y+7)&=847.$

we then do the prime factorization of 847, which i think is, $7*11*11$ . we have to find the numbers that multiply to 847 and then plug them into z+5, 3x=1,2y+7.

It has to be a positive, non-negative integer, right?

We also see that 3x+1=11 so we see that x=10/3 (which wont work).

So 3x+1=7, so x=2.

So 11 has to be in another term. It has to be in 2y+7=11 so y=2

for the last term we get z+5=11 so z=6

2+2+6=10

Hope this helps and if you want please consider giving me brainliest. :)

You might be interested in

If two events have probabilities which sum to 1, they are called ________ events.

Alenkasestr [34]

<span>An event with probability 0 never occurs. An event with probability 1 occurs on every trial</span>

8 0

3 years ago

Read 2 more answers

12. (x3y6)-2 + (x2y4)-3

Galina-37 [17]

Answer: 26xy-5

(x3y6)-2 + (x2y4)-3

(18xy)-2 + (8xy)-3

18xy -2 + 8xy - 3

26xy -5

7 0

3 years ago

What is the answer to (2x^2y)^3

natta225 [31]

Answer:

8x^6y^3
———————

6 0

3 years ago

For the first 80 miles of her road trip, Lina travels 10 miles/hour slower than she did for the remaining 50 miles of her trip.

Contact [7]

The expressions represent the total time of her trip in hours is $\rm \dfrac{80}{s} + \dfrac{50}{s+10}$ .

Given that,

For the first 80 miles of her road trip, Lina travels 10 miles/hour slower than she did for the remaining 50 miles of her trip.

It represents Lina's speed during the first part of her trip.

We have to determine,

Which expressions represent the total time of her trip in hours?

According to the question,

Let the speed traveled during the first part of the trip be "s".

The formula for calculating the distance is expressed as:

$\rm Distance = Speed \times Time$

For the first part of the journey:

Distance = 80 miles

Time = $\rm t_1$

And Speed = s

Substitute all the values in the formula,

$\rm Distance = Speed \times Time\\\\80 = s \times t_1\\\\ t_1= \dfrac{80}{s}$

For the second part of the trip;

Distance traveled = 50 miles

Speed = s + 10 (10miles/hour faster than the first part of the trip)

time = $\rm t_2$

Substitute all the values in the formula,

$\rm Distance = Speed \times Time\\\\50 = (s+10) \times t_2\\\\ t_2 = \dfrac{50}{s+10}$

Therefore,

Adding both the time to represent the total time of her trip in hours,

$\rm = t_1 +t_2 \\\\ = \dfrac{80}{s} + \dfrac{50}{s+10}$

Hence, The required expressions represent the total time of her trip in hours is $\rm \dfrac{80}{s} + \dfrac{50}{s+10}$ .

For more details refer to the link given below.

brainly.com/question/21043759

6 0

2 years ago

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

3 years ago