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asambeis [7]
3 years ago
9

-10 = -10 + 7m Thank you

Mathematics
2 answers:
My name is Ann [436]3 years ago
7 0

Step-by-step explanation:

-10= -10+7m

-10+10=7m

0=7m

m=<u>0</u>

<u> </u><u> </u><u> </u><u> </u><u> </u>7

m=0

hope it helps.

wariber [46]3 years ago
6 0
The answer is m=0
-10= -10+7m
+10 +10
0=7m
/7 /7 divide by 7 on both sides
m=0
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Area of a trapezium​
melamori03 [73]

Answer:

The area of a trapezium is computed with the following formula: Area = 1 2 × Sum of parallel sides × Distance between them.

Step-by-step explanation:

Hope it help! :)

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3 years ago
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32 divided by (12 - 4) + 7 write 2 steps then answer and i will give u brainliest!!
Artemon [7]

Answer:

12-4=8+7=15

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What are the x- and y- coordinates of point E, which partitions the directed line segment from A to B into a ratio of 1:2?
Solnce55 [7]

<u><em>The coordinates of the points A and B are not given. We'll assume: A(3,1) B(12,-5)</em></u>

Answer:

<em>The coordinates of E are (6,-1)</em>

Step-by-step explanation:

<u>Partition of a Segment</u>

The length of segment directed from A(xa,ya) to B(xb,yb) can be decomposed in its x,y coordinates:

x_{AB}=x_b-x_a

y_{AB}=y_b-y_a

If a point E is to be in the segment and partition it into a ratio 1:2, then

\displaystyle \frac{x_{AE}}{{x_{EB}}=\frac{y_{AE}}{{y_{EB}}=\frac{1}{2}

But

x_{AE}=x_e-x_a

y_{AE}=y_e-y_a

x_{EB}=x_b-x_e

y_{EB}=y_b-y_e

Then we set the equation:

\frac{x_{AE}}{{x_{EB}}=\frac{1}{2}

2(x_e-x_a)=x_b-x_e

Operating and rearranging

3x_e=x_b+2x_a

Solving

\displaystyle x_e=\frac{x_b+2x_a}{3}=\frac{12+6}{3}

x_e=6

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\displaystyle y_e=\frac{y_b+2y_a}{3}=\frac{-5+2}{3}

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The coordinates of E are (6,-1)

8 0
4 years ago
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Solve: what is 3p - 6 &gt; -6
RSB [31]

Answer:

0

Step-by-step explanation:

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1. Differentiate with respect to x:<br>​
natta225 [31]

9514 1404 393

Answer:

  a) y' = x^2(3x·ln(6x) +1)

  b) y' = 6e^(3x)/(1 -e^(3x))^2

Step-by-step explanation:

The applicable rules for derivatives include ...

  d(u^n)/dx = n·u^(n-1)·du/dx

  d(uv)/dx = (du/dx)v +u(dv/dx)

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__

(a)

  y=x^3\ln{(6x)}\\\\y'=3x^2\ln{(6x)}+\dfrac{x^3\cdot6}{6x}\\\\\boxed{\dfrac{dy}{dx}=3x^3\ln{(6x)}+x^2}

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(b)

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5 0
3 years ago
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