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Sergeu [11.5K]
3 years ago
13

Which set of equations has (2, 2) as its solution? (4 points) A and D B and D A and C B and C

Mathematics
1 answer:
Andrej [43]3 years ago
4 0
I don't think there are any equations that passes through (2,2). Please clarify
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Trigonometry please help
kotykmax [81]

Answer:

7.5

Step-by-step explanation:

x = 10 \tan \: 37 \degree \\  \\  \therefore \: x = 7.535540501 \\  \\ \therefore \: x  \approx 7.5

3 0
3 years ago
Read 2 more answers
Lan scores 5 out of 60 mark test.what is his score as a percentage to 1 decimal place
artcher [175]

Answer:

.1 approx.

Step-by-step explanation:

5/60 and x/100%

Cross multiplying...

5 x 100% = 5

5/60 ~= .1

( I dunno if I am right, so please check first.)

6 0
3 years ago
200/40+15 = ??<br><br><br>helpp mee ^^ ​
wel

Answer:20   ^v^ Please follow me

Step-by-step explanation:

200/40=5

5+15=20

5 0
3 years ago
Read 2 more answers
I have been stuck on the problem for so long. please help me:
Xelga [282]

Answer:

D-2

Step-by-step explanation:

You can use elimination to get the value of y. Move numbers to one side and x and ys to the other side. To use elimination, multiply the first equation with 2. You will get 2x+2y=24. add it with the equation below. y will be eliminated and you will get the value of x. X=10. Plug in x and get the y value. Y=2

6 0
3 years ago
Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100. The maximum value of f(x,y) is:
ryzh [129]

First find the critical points of <em>f</em> :

f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7

\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1

\dfrac{\partial f}{\partial y}=6y=0\implies y=0

so the point (1, 0) is the only critical point, at which we have

f(1,0)=-7

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t

Find the critical points of <em>g</em> :

\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0

\implies\sin t=0\text{ OR }1+5\cos t=0

\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi

where <em>n</em> is any integer. We get 4 critical points in the interval [0, 2π) at

t=0\implies f(10,0)=155

t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299

t=\pi\implies f(-10,0)=235

t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299

So <em>f</em> has a minimum of -7 and a maximum of 299.

4 0
3 years ago
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