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Jobisdone [24]
3 years ago
14

Need help on this please

Mathematics
2 answers:
Naily [24]3 years ago
6 0
There is no outlier. :)
FrozenT [24]3 years ago
5 0
No Outlier.

Hope this helps ;)
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If $x$, $y$, and $z$ are positive integers such that $6xyz+30xy+21xz+2yz+105x+10y+7z=812$, find $x+y+z$.
harina [27]

Answer:

10

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. :)

8 0
4 years ago
Kelsey wants to buy a new video game.
Ahat [919]

Answer: c: $65

Step-by-step explanation:

3 0
3 years ago
What shape is a quadrilateral with two sets of parallel equal sides and right angles
Nostrana [21]
A square because it has parrel sides and has equal sides and also has right angles
8 0
3 years ago
Find the value of w in the diagram below.<br> w<br> 3<br> 8
sleet_krkn [62]
What diagram are you talking about because I don’t see one
3 0
3 years ago
Find the average value of the function y = 6 - x2 over the interval [-1, 4]
kari74 [83]

Answer:

The average value of the function f(x) = 6 - x^{2} over the interval [-1,4] is \frac{5}{3}.

Step-by-step explanation:

The average value of a function over an interval is represented by this integral:

\bar y = \frac{1}{b-a}\cdot \int\limits^{b}_{a} {f(x)} \, dx

Where:

a, b - Lower and upper bounds of the interval, dimensionless.

f(x) - Function, dimensionless.

If a = -1, b = 4 and f(x) = 6 - x^{2}, the average value of the function is:

\bar y = \frac{1}{4-(-1)}\int\limits^{4}_{-1} {6-x^{2}} \, dx

\bar y = \frac{6}{5}\int\limits^{4}_{-1} \, dx - \frac{1}{5}\int\limits^{4}_{-1} {x^{2}} \, dx

\bar y = \frac{6}{5}\cdot x |_{-1}^{4} - \frac{1}{15}\cdot x^{3}|_{-1}^{4}

\bar y = \frac{6}{5}\cdot [4-(-1)]- \frac{1}{15}\cdot [4^{3}-(-1)^{3}]

\bar y = \frac{5}{3}

The average value of the function f(x) = 6 - x^{2} over the interval [-1,4] is \frac{5}{3}.

4 0
3 years ago
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