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docker41 [41]
3 years ago
13

What is the solution for the system of equations {2x-3y-2z=-7, x-3y-3z=4, x-2y-z=-3

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
7 0
A suitable matrix calculation tells you (x, y, z) = (-8, -1, -3).

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Help pls! this is due tonight. thanks!
erma4kov [3.2K]

Answer:

Neon: 0

Oxide:  -2

Copper:   2

Tin:  0

Step-by-step explanation:     To solve you must add/subtract the numbers to get the charge, for example if its -30 charge from electrons and 10 charge from protons, -30 + 10 = -20     therefore the charge is -20

6 0
3 years ago
Read 2 more answers
You bike 1 mile the first day of your​ training, 1.3 miles the second​ day, 1.9 miles the third​ day, and 3.1 miles the fourth d
JulsSmile [24]

Answer:

19.9 miles

Step-by-step explanation:

In this problem we have:

d_1=1 mi is the distance travelled during the 1st day

d_2=1.3 mi is the distance travelled during the 2nd day

d_3=1.9 mi is the distance travelled during the 3rd day

d_4=3.1 mi is the distance travelled during the 4th day

We notice that the difference between the distance travelled on the (n+1)-th day and the distance travelled on the n-th day doubles every day. In fact:

d_2-d_1=0.3\\d_3-d_2=2\cdot 0.3 = 0.6\\d_4-d_3=2\cdot 0.6 = 1.2

Which can be rewritten using the general formula:

d_{n+1}-d_n=2(d_n-d_{n-1})

This means that

d_{n+1}=d_n+2(d_n-d_{n-1})

By applying this formula recursively, we can find the 7th term, which is the distance travelled on the 7th day:

d_1=1\\d_2=1.3\\d_3=1.9\\d_4=3.1\\d_5=3.1+2\cdot 1.2=5.5\\d_6=5.5+2\cdot 2.4=10.3\\d_7=10.3+2\cdot 4.8=19.9 mi

So, the  distance travelled on the 7th day is 19.9 miles.

7 0
3 years ago
What times 4will get to 80
Makovka662 [10]
20 times 4 will get to 80
7 0
3 years ago
Read 2 more answers
Y=2(x+1)^2 has how many real roots
oee [108]

The equation y= 2(x+1)^2 has one real root and that is x=-1.

What is real roots of the equation?

    We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation X^2-7x+12=0 is solved, the actual roots are 3 and 4.

Here given,

=> y = 2(x+1)^2

Take y=0 then,

=> 2(x+1)^2=0

=> (x+1)^2=0

=>(x+1)=0

=> x=-1

Hence the given equation has one real root and that is x=-1.

To learn more about real roots refer the below link

brainly.com/question/24147137

#SPJ1

8 0
1 year ago
Read 2 more answers
What is the value of x in the equation (StartFraction one-half EndFractionx + 12) = StartFraction one-half EndFraction(StartFrac
anastassius [24]

Answer:

x =-24

Step-by-step explanation:

Given

(\frac{2}{3})(\frac{1}{2}x + 12) = (\frac{1}{2})(\frac{1}{3}x + 14) - 3

Required

Solve for x

(\frac{2}{3})(\frac{1}{2}x + 12) = (\frac{1}{2})(\frac{1}{3}x + 14) - 3

Open all brackets

\frac{2}{3}*\frac{1}{2}x + \frac{2}{3}*12 = \frac{1}{2}*\frac{1}{3}x + \frac{1}{2}*14 - 3

\frac{2 * 1}{3 *2}x + \frac{2 * 12}{3}= \frac{1 * 1}{2 * 3}x + \frac{1 * 14}{2} - 3

\frac{1}{3}x + \frac{24}{3}= \frac{1}{6}x + \frac{14}{2} - 3

\frac{1}{3}x +8= \frac{1}{6}x + 7 - 3

Collect like terms

\frac{1}{3}x - \frac{1}{6}x =7 - 3 -8

\frac{1}{3}x - \frac{1}{6}x =-4

Solve fraction

\frac{2-1}{6}x =-4

\frac{1}{6}x =-4

Multiply both sides by 6

6 * \frac{1}{6}x =-4 * 6

x =-4 * 6

x =-24

8 0
3 years ago
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