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Thepotemich [5.8K]
3 years ago
14

Is this right? i’m confused.

Mathematics
2 answers:
lora16 [44]3 years ago
8 0
Yes that is correct
Ymorist [56]3 years ago
3 0

Answer:

Yes you are correct

Step-by-step explanation:

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Can anyone explain how this problem is done ?
Svetach [21]

B

Step-by-step explanation:

6 0
3 years ago
Solve for p: m= 8 -2(p-m)
makkiz [27]
Solve for p by simplifying both sides of the equation, then isolation the variable
p = 1 + m/2

Hope this helps! :)
4 0
3 years ago
Read 2 more answers
Use a matrix to solve the system:
Romashka-Z-Leto [24]

Answer:

(2.83 , 1 , 4)

Step-by-step explanation:

2x+2y-z=4\\4x-2y-2z=2\\3x+3y-4z=-4\\

Rewrite these equations in matrix form

\left[\begin{array}{ccc}2&2&-1\\4&-2&-2\\3&3&-4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\

we can write it like this,

AX=B\\X=A^{-1}B

so to solve it we need to take the inverse of the 3 x 3 matrix A then multiply it by B.

We get the inverse of matrix A,

A^{-1}=\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right]  \\

now multiply the matrix with B

X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right]\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2.83\\1\\4\end{array}\right] \\

4 0
3 years ago
I don't understand this.
sladkih [1.3K]
You mean the word or number
7 0
3 years ago
Match the equations with their solutions over the interval [0, 2π].
nignag [31]

I solved this using a scientific calculator and in radians mode since the given x's is between 0 to 2π. After substitution, the correct pairs are:

cos(x)tan(x) – ½ = 0 → π/6 and 5π/6

cos(π/6)tan(π/6) – ½ = 0

cos(5π/6)tan(5π/6) – ½ = 0

 

sec(x)cot(x) + 2 = 0 → 7π/6 and 11π/6

sec(7π/6)cot(7π/6) + 2 = 0

sec(11π/6)cot(11π/6) + 2 = 0

 

sin(x)cot(x) + 1/sqrt2 = 0 → 3π/4 and 5π/4

sin(3π/4)cot(3π/4) + 1/sqrt2 = 0

sin(5π/4)cot(5π/4) + 1/sqrt2 = 0

 

csc(x)tan(x) – 2 = 0 → π/3 and 5π/3

csc(π/3)tan(π/3) – 2 = 0

csc(5π/3)tan(5π/3) – 2 = 0

5 0
4 years ago
Read 2 more answers
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