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Rudik [331]
3 years ago
13

Match each graph to the equation of its line.

Mathematics
2 answers:
miskamm [114]3 years ago
4 0

Answer:

the first graph is y=4x-4

the second is y=x+2

the third is y=-2x-4

the fourth is y=-x+2

Step-by-step explanation:

please rate and mark brainliest if correct :>

Diano4ka-milaya [45]3 years ago
3 0
What the guy above is correct
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2 years ago
If a = t and b=2 then find the value of - 2ab and - 3a+b
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Step-by-step explanation:

Given,

a=t

b=2

-2ab=-2×t×2 = -4t

-3a+b=-3×t+2= -3t+2

4 0
3 years ago
A sequence is defined by the formula f(n+1)=f(n)-3. If f(4)=22, what is f(1)?
Leto [7]

answer : f(1) is 31

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4 0
3 years ago
Which value for x would make the perimeter of this rectangle equal 48 inches
RSB [31]

Answer:

L = 16; W=8: are the correct values.

Step-by-step explanation:

Start off with 2 math statements:

First, for the perimeter:  2W x 2L = 48

Then, for the area:  W x L = 128

Next, we want to take the 2nd equation; then solve it for "W"

W x L = 128

Divide out both sides by "L", to get the "W" equals:

W = 128/L

Next, go back to the first equation.  Where it indicates "W", put in the value of:  128/L in its place; to get:

2L + 2*(128/L) = 48; which gives us:

2L + 256/L = 48

Multiply everything by "L", to get:

2L2 + 256 = 48L

Next, divide out everything by 2, to get:

L2 + 128 = 24L

Next, move the 24L over to the left, to get:

L2 - 24L + 128 = 0

(L-16)* (L-8) = 0

L-16 = 0; L-8 = 0

Which gives us:  L=16; L=8

Let's go with the L=16 first; back into the original equation:

f(16) = 2*16 + 2W = 48

32 + 2W = 48

-=32         -32

2W = 16

Divide out both sides by 2, to get:

W = 8

Next, test our values, by plugging them back into both original equations:

2*16 + 2*8 = 48

32 + 16 = 48        [Check]

L x W = 128

16 * 8 = 128

128 = 128          [Check]

5 0
2 years ago
I’m in Algebra 2 and we’re currently on the subject of “Solving Absolute Value Equations” and I’m having trouble with this equat
saw5 [17]

let's bear in mind that an absolute value expression is in effect a piece-wise expression,  namely it has a ± versions of the same expression.

\bf 5|3x-4| = x+1\implies |3x-4|=\cfrac{x+1}{5}\implies \begin{cases} +(3x-4)=\cfrac{x+1}{5}\\[1em] -(3x-4)=\cfrac{x+1}{5} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(3x-4)=\cfrac{x+1}{5}\implies 3x-4=\cfrac{x+1}{5}\implies 15x-20=x+1 \\\\\\ 14x-20=1\implies 14x=21\implies x = \cfrac{21}{14}\implies \boxed{x=\cfrac{3}{2}} \\\\[-0.35em] ~\dotfill\\\\ -(3x-4)=\cfrac{x+1}{5}\implies -3x+4=\cfrac{x+1}{5}\implies -15x+20=x+1 \\\\\\ 20=16x+1\implies 19=16x\implies \boxed{\cfrac{19}{16}=x}

6 0
3 years ago
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