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MissTica
3 years ago
11

Write two mixed numbers that have a sum of 3.

Mathematics
1 answer:
Lina20 [59]3 years ago
7 0
1 1/2+ 1 1/2 = 3 .........
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What is the missing statement in step 10 of the proof?
telo118 [61]

Answer:

c/sin C = b/sin C

Step-by-step explanation:

Look at the statement in the previous step and the reason in this step.

c sin B = b sin C

Divide both sides by sin B sin C:

(c <em>sin B</em>)/(<em>sin B</em> sin C) = (b <em>sin C</em>)/(sin B <em>sin C</em>)

c/sin C = b/sin B

7 0
3 years ago
USe Diagram to solve
BlackZzzverrR [31]

Answer:

Answer is 12, D

6 0
3 years ago
Simplify the expression
Molodets [167]

Answer:

8x^8/3 y^4 - The first option

Step-by-step explanation:

The first thing we need to do is the exponent outside the bracket and leave the 8 till last, because exponents always come before multiplying coefficients. When an term with an exponent is multiplied by another exponent outside a bracket, the exponents of both terms are multiplied by the exponent outside the bracket.

This means that the expression is now:

8(x^2*4/3 y^3*4/3)

First we can so the x term. The x term already has an exponent of 2, so the 2 is multiplied by the 4/3 exponent outside the bracket. 2*4/3 = 8/3, so the x term is now: x^8/3

The same happens to the y term: 3*4/3 simplifies to 4, so the y term is now y^4.

So now our expression is:

8(x^8/3 y^4)

Now the 8 outside the bracket simply multiplies on to the whole term so we finish with:

8x^8/3 y^4 - The first option.

Hope this helped!

7 0
3 years ago
What is the equation in point-slope form for the line parallel to y = –2x + 10 that contains J(6, 8)?
Liula [17]
Y-y1= m(x1-x2)
using this formula
the answer is d
because slope has to remain the same if its parallel

5 0
3 years ago
Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81
Anton [14]
Using Lagrange multipliers, we have the Lagrangian

L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)

with partial derivatives (set equal to 0)

L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}
L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}
L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}
L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81

Substituting the first three equations into the fourth allows us to solve for \lambda:

x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}

For each possible value of \lambda, we get two corresponding critical points at (\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3).

At these points, respectively, we get a maximum value of f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3 and a minimum value of f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3.
5 0
3 years ago
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