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JulijaS [17]
2 years ago
13

System with 3 variable. solve pleaseeee x+4y-5z=-7 3x+2y+3z=7 2x+y+5z=8

Mathematics
1 answer:
7nadin3 [17]2 years ago
6 0
To solve this we need to form 2 equations with only 2 variables.
x+4y-5z=-7
2x+y+5z=8
3x+5y=1

15x+10y+15z=35
6x+3y+15z=24
9x+7y=11

Now we need to use these to find x and y
9x+15y=3
9x+7y=11
8y=-8
y=-1
Therefore:
x=2
From this we can find that:
2(2)-1+5z=8
Therefore z=1
The answer is x=2, y=-1, z=1
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Which of the following is the complex conjugate of the complex number below? 4+51i
Fynjy0 [20]
Answer is C

The conjugate of the complex number a+bi is a-bi

you just reverse the sign of the imaginary part of the number ; so it is 4-51i

5 0
3 years ago
The number of patients treated at Dr. McCormick’s dentist office each day was recorded for nine days. These are the data: 13, 16
arlik [135]
Hi!

Let's start with the mean.

The mean is all the numbers added together, divided by how many numbers there are. 

13+16+16+10+6+12+4+12+16=105
there are 9 numbers so..
105/9 = 11.7

The mean is 11.7 

Find the median. 

The median is the number in the middle of the data set when the numbers are in numerical order.

4, 6, 10, 12, 12, 13, 16, 16, 16

The median is 12. 

Find the mode.

The mode is the number that is repeated the most.

4, 6, 10, 12, 12, 13, 16, 16, 16

The mode is 16

Find the range.

The range is the difference of the smallest number from the biggest number.

16 - 4 = 12

The range is 12

The answer is 11.7, 12, 16, 12 

Hope this helps! :)
4 0
3 years ago
5/12=10/? Someone please help
nasty-shy [4]

Answer:

0.4167

Step-by-step explanation:

hope i help

4 0
2 years ago
Read 2 more answers
The question is attached in the picture below. (University level)
Luba_88 [7]

Answer:

nope

Step-by-step explanation:

Apply dot product which must = 0 for orthogonality.

In this case,

-(3)(2) + (-2)(-2) + (-3)(1) + (1)(4) = -1.

4 0
2 years ago
Differential cos^2x dy/dx =e^y-tanx​
gulaghasi [49]

Answer:

y=t−1+ce

−t

where t=tanx.

Given, cos

2

x

dx

dy

+y=tanx

⇒

dx

dy

+ysec

2

x=tanxsec

2

x ....(1)

Here P=sec

2

x⇒∫PdP=∫sec

2

xdx=tanx

∴I.F.=e

tanx

Multiplying (1) by I.F. we get

e

tanx

dx

dy

+e

tanx

ysec

2

x=e

tanx

tanxsec

2

x

Integrating both sides, we get

ye

tanx

=∫e

tanx

.tanxsec

2

xdx

Put tanx=t⇒sec

2

xdx=dt

∴ye

t

=∫te

t

dt=e

t

(t−1)+c

⇒y=t−1+ce

−t

where t=tanx

8 0
2 years ago
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