1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Sliva [168]
3 years ago
12

1 4/5 x 5/12 another question pls urgent​

Mathematics
1 answer:
gogolik [260]3 years ago
4 0

Answer:

45/60 or 3/4

Step-by-step explanation:

You have to convert any mixed number into a fraction

1 4/5 = 9/5

5/12 = 5/12

Then you multiply across

9 x 5 = 45

5 x 12 =60

Then you would simplify the answer

<h2>\frac{45}{60} = \frac{3}{4}</h2>
You might be interested in
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
Find a vector function, r(t), that represents the curve of intersection of the two surfaces. the paraboloid z = 7x2 + y2 and the
Akimi4 [234]
Letting x=t, we get y=3x^2=3t^2 and z=7x^2+y^2=7t^2+(3t^2)^2=7t^2+9t^4, so we can parameterize the intersection by

\mathbf r(t)=\langle t,3t^2,7t^2+9t^4\rangle

where -\infty.

Image attached.

5 0
3 years ago
If both cars in Exercise 62 are on one side of the plane and if the angle of depression to one car is 38∘ and that to the other
Sati [7]

The two cars lying on one side of the plane are 50 meters apart.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

let us assume that the height is 100 m. Let d represent the horizontal distance. For depression of 38°, Ф = 90 - 38 = 52°, hence:

tan(52°) = d / 100

d = 128 m

For depression of 52°, Ф = 90 - 52 = 38°, hence:

tan(38°) = d / 100

d = 78 m

Distance apart = 128 - 78 = 50 m.

The two cars lying on one side of the plane are 50 meters apart.

Find out more on equation at: brainly.com/question/2972832

#SPJ1

7 0
2 years ago
3/4 x 1/2 <br> Help need it plss ;-;
Shkiper50 [21]

Answer:

3/8

Step-by-step explanation:

3/4 divide 2 = 4÷2= 3/8

4 0
3 years ago
Read 2 more answers
-4|x-11|=-16 solve for x in the equation
BigorU [14]

Answer:

x = 7 and 15

Step-by-step explanation:

Divide both sides by -4...

| x - 11 | = 4

because

| -4 | = 4 and

| 4 | = 4

x - 11 = -4 and x - 11 = 4

solve for x in both equations.

x = 7 and x = 15

5 0
3 years ago
Other questions:
  • Find an equation in slope intercept for the line 3x-4y=7
    9·1 answer
  • Given the function f(x) = 4(2)x, Section A is from x = 1 to x = 2 and Section B is from x = 3 to x = 4.
    9·1 answer
  • Write an equation in slope intercep form that passes thru the points (10,2) and (2,-2)
    14·1 answer
  • S squared - 16s - 57
    7·1 answer
  • Mastery challenge
    11·1 answer
  • Calories consumed by members of a track team the day before a race are normally distributed, with a mean of 1,800 calories and a
    8·1 answer
  • Help I need this done fats!
    9·2 answers
  • I've been stuck on this for a while now​
    12·2 answers
  • Thad is Mac d a d d y
    14·2 answers
  • Select the correct days. Gina tracks the low temperatures for a city over six days. Identify the days colder than -1.75°F. I WIL
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!