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Doss [256]
3 years ago
7

A = 1/2h(m + n) for n​

Mathematics
1 answer:
nasty-shy [4]3 years ago
5 0

The answer is n = 2A/h

Solve for n:

Multiply both sides by 2.

Divide both sides by h.

Finally, subtract m from both sides.

Answer is: n = 2A/h

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What is the equation in slope intercept form for (2,5) (3,12)
kondaur [170]

Slope formula: y2-y1/x2-x1

= 12-5/3-2

= 7/1

= 7

_____

Best Regards,

Wolfyy :)

7 0
3 years ago
Solve the differential. This was in the 2016 VCE Specialist Maths Paper 1 and i'm a bit stuck
Nimfa-mama [501]
\sqrt{2 - x^{2}} \cdot \frac{dy}{dx} = \frac{1}{2 - y}
\frac{dy}{dx} = \frac{1}{(2 - y)\sqrt{2 - x^{2}}}

Now, isolate the variables, so you can integrate.
(2 - y)dy = \frac{dx}{\sqrt{2 - x^{2}}}
\int (2 - y)\,dy = \int\frac{dx}{\sqrt{2 - x^{2}}}
2y - \frac{y^{2}}{2} = sin^{-1}\frac{x}{\sqrt{2}} + \frac{1}{2}C


4y - y^{2} = 2sin^{-1}\frac{x}{\sqrt{2}} + C
y^{2} - 4y = -2sin^{-1}\frac{x}{\sqrt{2}} - C
(y - 2)^{2} - 4 = -2sin^{-1}\frac{x}{\sqrt{2}} - C
(y - 2)^{2} = 4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C


y - 2 = \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}
y = 2 \pm\sqrt{4 - 2sin^{-1}\frac{x}{\sqrt{2}} - C}

At x = 1, y = 0.
0 = 2 \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}
-2 = \pm\sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}

4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C > 0
\therefore 2 = \sqrt{4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C}


4 = 4 - 2sin^{-1}\frac{1}{\sqrt{2}} - C
0 = -2sin^{-1}\frac{1}{\sqrt{2}} - C
C = -2sin^{-1}\frac{1}{\sqrt{2}} = -2\frac{\pi}{4}
C = -\frac{\pi}{2}

\therefore y = 2 - \sqrt{4 + \frac{\pi}{2} - 2sin^{-1}\frac{x}{\sqrt{2}}}
6 0
3 years ago
What is the Domain and Range *DON'T FORGET PARENTHESES AND BRACKETS*
Alborosie

Answer:

D= {-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6}

R={-6,-5,-4,-3,-2,-1,1,2,5)

Step-by-step explanation:

5 0
2 years ago
Mr.Ward runs a lot. He ran 45 minutes each day, 5 days each week , for 16 weeks . In that time, he ran 450 miles. What was his a
Savatey [412]

Answer:

7.5 miles per hour.

Step-by-step explanation:

We have been given that Mr. Ward runs a lot. He ran 45 minutes each day, 5 days each week, for 16 weeks.

First of all, we will find time for that Mr. Ward ran in 16 weeks.

We will multiply 5 by 16 to find number of days for that Mr. Ward ran and then we will multiply the result by 45 minutes to find the time.

\text{Time for that Mr. Ward ran in 16 weeks}=16\times 5\times 45\text{ minutes}

\text{Time for that Mr. Ward ran in 16 weeks}=3600\text{ minutes}

Now, we will divide 3600 minutes by 60 minutes to convert time into hours as:

\text{Hours}=\frac{3600}{60}=60

Now, we will divide 450 miles by 60 hours to find Mr. Ward's average speed as:

\text{Mr. Ward's average speed}=\frac{450\text{ miles}}{\text{60 hours}}

\text{Mr. Ward's average speed}=\frac{7.5\text{ miles}}{\text{ Hour}}

Therefore, Mr. Ward's average speed in 7.5 miles per hour.

7 0
3 years ago
1) Find the mean of the following numbers: 5, 11, 8, 8, 7, 4, 10, 9, 7, 7, 6​
SashulF [63]
Mean = 5+11+8+8+7+4+10+9+7+7+6 divided by the number of observations which is 10

Therefore, mean = 73/10
Mean = 7.3
4 0
3 years ago
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