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zlopas [31]
3 years ago
7

What is the slope - intercept form equation of the line graphed below ?

Mathematics
1 answer:
Jlenok [28]3 years ago
7 0
Slope = (7-2)/(3-0) = 5/3

slope intercept b = 2

equation y= 5/3x +b

<span>answer: B. y=5/3x + 2 </span>
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Please help. I’ll mark you as brainliest if correct!
balu736 [363]

Answer: x= -1, z=2, y= -4

Step-by-step explanation:

System of equations:

-5x - 4y - 3z= 15  +

<u>-10x + 4y + 6z= 6</u>

-15x         + 3z = 21  ------>  3 (-5x + z) = 7.3

-5x + z = 7

now,

-10x + 4y + 6z= 6

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

14 + 4y + 4z = 6

7 + 2y + 2z = 3

2y + 2z= -4

y+z=-2

Now we were using the equation: 20x + 4y + 4z = -28

20x + 4(y+z) = 20x -8= - 28

20 x = -20

x= -1

With this we can find y and z

X=-1

-5x + z = 7

z= 2

y+z=-2

y=-4

Finally we have: x= -1, z=2, y= -4

I hope this can help you.

Thank you

3 0
3 years ago
3÷ 1/2 1÷1/4 1/2÷2 1/3=4 2÷ 1/6 1/4÷3​
Pani-rosa [81]

Answer: [45]

Step-by-step explanation:

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4 0
3 years ago
Please help me with this homework
Vadim26 [7]

Answer:

s=Q/Rt - 1/t

Step-by-step explanation:

5 0
2 years ago
What 2 numbers should be used to find 231 - 198 using the 10 method for subtracting quickly in your head?
Airida [17]

Answer: "Count 2 and then count 31 more."

Step-by-step explanation:

We have the equation:

231 - 198

Now, the negative number is kinda ugly, so i will write it as:

200 - 2 = 198

Is a lot easier work with 200 and 2, than with 198.

Then the equation is now:

231 - (200 - 2)

And the left number we also have a "200", so it can be written as:

200 + 31 = 231

As this is a sum, we can ignore the parentheses.

200 + 31 - 200 + 2

31 + 2

Then the correct option is:

"Count 2 and then count 31 more."

4 0
3 years ago
Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7
rosijanka [135]
\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx

Notice that x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1). Decompose the integrand into partial fractions:

\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}
=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}

Integrating term-by-term, you get

\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx
=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C
5 0
3 years ago
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