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Lyrx [107]
4 years ago
15

PLEASE ANSWER ASAP!

Mathematics
1 answer:
Likurg_2 [28]4 years ago
8 0

Answer:

d.- 6m - n + 3

Step-by-step explanation:

-3(2m-1)-n

-6m + 3 -n

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Math question picture attached
babunello [35]

To find the slope, you use the slope formula (m):

m=\frac{y_2-y_1}{x_2-x_1}    and plug in the two points

(-3, 4) = (x₁, y₁)

(2, -1) = (x₂, y₂)

m=\frac{y_2-y_1}{x_2-x_1}

m=\frac{-1-4}{2-(-3)}       [two negatives cancel each other out and become positive]

m=\frac{-1-4}{2+3}

m=\frac{-5}{5}

m = -1        The slope is -1

4 0
3 years ago
Read 2 more answers
If you know the answer for this please help me I would really appreciate it thank you!
STALIN [3.7K]

Answer:

  line QR

Step-by-step explanation:

Plane QRB is the front face. Plane TSR is the top face. The planes intersect at the top front edge, line QR.

3 0
3 years ago
What are the coordinates to this?
Hoochie [10]

Answer:

D (-5,-2) E (0,-2), and F (-4,-6)

Step-by-step explanation:

hope it helps

4 0
3 years ago
Read 2 more answers
This box can be packed with 48 units cubes. The edge length of each unit cube is 1 meter. What is the volume of the box?
Dafna1 [17]

48 m³

Step-by-step explanation:

A cube has equal length width and height

A cube in this case has length 1m meaning the width and height are also 1 m each in length.

The volume of the cube is (L * W * H) =  1 * 1 * 1 = 1 m³

If 48 of these can fit in the bigger box, then the bigger box has the volume of;

48 * 1 m³ = 48 m³

Learn More:

For more on volume of a cube check out;

brainly.com/question/7543014

brainly.com/question/9719931

#LearnWithBraily

8 0
3 years ago
Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8​
Ivanshal [37]

It looks like the differential equation is

\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}

Factorize the right side by grouping.

xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)

xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)

Now we can separate variables as

\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx

Integrate both sides.

\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx

\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx

\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}

You could go on to solve for y explicitly as a function of x, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0
2 years ago
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