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lukranit [14]
3 years ago
8

BTW this is Algebra 2

Mathematics
1 answer:
djverab [1.8K]3 years ago
5 0

<h3>1) <u>What is the solution for the system of equations</u></h3>

-2x+y=2...(i)

y=4x-1...(ii)

substitute y in (i)

-2x+(4x-1)=2

-2x+4x-1=2

-2x+4x=2+1

2x=3

2x/2=3/2

x=3/2

replacing x in (ii)

y=4(3/2)-1

y=6-1

y=5

solution=(3/2,5)

<h3 /><h3>2) <u>Solve the system of equations.</u></h3>

x-y+z=6 ...(i)

4y+1=z ...(ii)

3y=2z...(iii)

from(ii),

z=4y+1

substitute z in (iii)

3y=2(4y+1)

3y=8y+2

3y-8y=2

-5y=2

-5y/5=2/5

y=-2/5

substitute y in (ii)

z=4(-2/5)+1

z=-3/5

substitute z and y in (i)

x-y+z=6

x=6+y-z

x=6+(-2/5)-(-3/5)

x=31/5

solution=(31/5,-2/5,-3/5)

<h3>3) <u>What is the solution for the system of equations</u></h3>

-4x+y=-2 ...(i)

y+5=-6x...(ii)

from (i),

y=-2+4x

substitute y in (ii)

(-2+4x)+5=-6x

-2+4x+5=-6x

-2+5=-6x-4x

3=-10x

3/-10=-10x/-10

x=-3/10

substitute x in;;y=-2+4x

y=-2+4(-3/10)

y=-2-6/5

y=-16/5

solution=(-3/10,-16/5)

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morpeh [17]
(2b^4)^(-1)=1/(2b^4)
 
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8 0
3 years ago
Solve the inequality below.<br><br> f2−9&gt;−5
Salsk061 [2.6K]

Answer:

f>2  or f<-2

Step-by-step explanation:

f^2−9>−5

Add 9 to each side

f^2−9+9>−5+9

f^2 >4

Take the square root of each side, remembering to flip the sign of  the inequality for the negative

f>2  or f<-2

5 0
3 years ago
Read 2 more answers
Question 16 of 17
FrozenT [24]
The answer is B. It is a relation and also a function at the same time because it satisfies a specific domain
3 0
3 years ago
Need help thanksssssssss
Archy [21]

Answer:

Volume: 112 m³.

Surface area: 172 m².

Step-by-step explanation:

The volume is the base times height times length. So, the volume will be 2 * 8 * 7 = 16 * 7 = 112 m³.

The surface area is 2lw + 2lh + 2wh. l = 8; w = 7; h = 2.

2(8)(7) + 2(8)(2) + 2(7)(2) = 2 * 56 + 2 * 16 + 2 * 14 = 112 + 32 + 28 = 112 + 60 = 172 m².

Hope this helps!

4 0
3 years ago
Help me please
kari74 [83]

Answer:

\frac{ \cos(80) }{ \sin(10) }  +  \frac{ \sin(20) }{ \cos(70) }  = 2 \\  \frac{ \cos(90 - 10) }{ \sin(10) }  +  \frac{ \sin(20) }{ \cos(9 0 - 20) }  = 2 \\  \frac{ \sin(10) }{ \sin(10) }  +  \frac{ \sin(20) }{ \sin(20) }  = 2 \\ 1 + 1 = 2 \\ 2 = 2 \\  \\  \frac{ \cot(40) }{ \tan(50) }  +  \frac{ \cos(65) }{ \sin(115) }  = 2 \\  \frac{ \cot(90 - 50) }{ \tan(50) }  +  \frac{ \cos(65) }{ \sin(90 + 65) }  = 2 \\  \frac{ \tan(50) }{ \tan(50) }  +  \frac{ \cos(65) }{ \cos(65) }  = 2 \\ 1 + 1 \\  = 2

6 0
3 years ago
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