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Pie
3 years ago
12

What is the slope of the line described by this equation? ​ 3y+2x=12 ​

Mathematics
1 answer:
Len [333]3 years ago
8 0
<span>-2/3 is the awnser hope it helps fam</span>
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11. 3/4 of 12=<br> 12. 6/7 of 49<br> 15. 2/3x27<br> 16. 3/8 x 16
Zarrin [17]

Step-by-step explanation:

3/4 of 12

3/4 × 12 = 9

6/7 of 49

6/7 × 49 = 42

2/3 × 27 = 18

3/8 x 16 = 6

8 0
3 years ago
What is the equation of the horizontal line through (1,9)
stealth61 [152]

Answer:

slope=9

Step-by-step explanation:

the slope is 9, because you do the y/x or rise over run, so you do for every 9 up 1 right to get a slope of 9/1 or 9

3 0
3 years ago
Read 2 more answers
Find the absolute maximum and minimum values of f(x, y) = x+y+ p 1 − x 2 − y 2 on the quarter disc {(x, y) | x ≥ 0, y ≥ 0, x2 +
Andreas93 [3]

Answer:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

Step-by-step explanation:

In order to find the absolute max and min, we need to analyse the region inside the quarter disc and the region at the limit of the disc:

<u>Region inside the quarter disc:</u>

There could be Minimums and Maximums, if:

∇f(x,y)=(0,0) (gradient)

we develop:

(1-2x, 1-2y)=(0,0)

x=1/2

y=1/2

Critic point P(1/2,1/2) is inside the quarter disc.

f(P)=1/2+1/2+p1-1/4-1/4=1/2+p1

f(0,0)=p1

We see that:

f(P)>f(0,0), then P(1/2,1/2) is a maximum relative

<u>Region at the limit of the disc:</u>

We use the Method of Lagrange Multipliers, when we need to find a max o min from a f(x,y) subject to a constraint g(x,y); g(x,y)=K (constant). In our case the constraint are the curves of the quarter disc:

g1(x, y)=x^2+y^2=1

g2(x, y)=x=0

g3(x, y)=y=0

We can obtain the critical points (maximums and minimums) subject to the constraint by solving the system of equations:

∇f(x,y)=λ∇g(x,y) ; (gradient)

g(x,y)=K

<u>Analyse in g2:</u>

x=0;

1-2y=0;

y=1/2

Q(0,1/2) critical point

f(Q)=1/4+p1

We do the same reflexion as for P. Q is a maximum relative

<u>Analyse in g3:</u>

y=0;

1-2x=0;

x=1/2

R(1/2,0) critical point

f(R)=1/4+p1

We do the same reflexion as for P. R is a maximum relative

<u>Analyse in g1:</u>

(1-2x, 1-2y)=λ(2x,2y)

x^2+y^2=1

Developing:

x=1/(2λ+2)

y=1/(2λ+2)

x^2+y^2=1

So:

(1/(2λ+2))^2+(1/(2λ+2))^2=1

\lambda_{1}=\sqrt{1/2}*-1 =-0.29

\lambda_{2}=-\sqrt{1/2}*-1 =-1.71

\lambda_{2} give us (x,y) values negatives, outside the region, so we do not take it in account

For \lambda_{1}: S(x,y)=(0.70, 070)

and

f(S)=0.70+0.70+p1-0.70^2-0.70^2=0.42+p1

We do the same reflexion as for P. S is a maximum relative

<u>Points limits between g1, g2 y g3</u>

we need also to analyse the points limits between g1, g2 y g3, that means U(0,0), V(1,0), W(0,1)

f(U)=p1

f(V)=p1

f(W)=p1

We can see that this 3 points are minimums relatives.

<u>Conclusion:</u>

We compare all the critical points P,Q,R,S,T,U,V,W an their respective values f(x,y). We find that:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

4 0
3 years ago
Suppose a triangle has sides a, b, and c, and the angle opposite the side of length b is obtuse. What must be true?
anyanavicka [17]

Answer: D.a^2 + b^2 < c^2

Step-by-step explanation:

In the given figure, we have an obtuse triangle which has sides a, b, and c, and the angle opposite the side of length b is obtuse.

∵ In a triangle, the side opposite to the largest angle is largest.

Thus, the largest side in then given obtuse triangle= c

The Pythagorean inequalities theorem says that If a triangle is obtuse than the square of the largest side is greater than the sum of square of other two sides.

Therefore, we have

a^2 + b^2 < c^2

5 0
3 years ago
I have 5 digits. My 9 is worth 9 ∗ 10,000. My 2 is worth 2 thousand. One of my 7s is worth 70. The other is worth 10 times as mu
Natasha2012 [34]
The answer is 92,776 because 10 times as much as 70 is 700 which gives the answer for the hundreds position.
6 0
3 years ago
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