All categories

4 years ago

How do you do 6/3 (improper fraction) on a number line?

Mathematics

1 answer:

ss7ja [257]4 years ago

8 0

Answer:

The number ins't simplified

Step-by-step explanation:

If you convert it into a mixed number it will be a lot easier to see where it goes. In this case tho 6 divided by 3 just equals 2, so the two is what you put on a number line.

You might be interested in

Lets see who is smart lol

Simora [160]

9514 1404 393

Answer:

$\square\quad{y=\dfrac{1}{2}x+4}\\\\\square\quad{\dfrac{y-5}{x-2}=\dfrac{1}{2}}$

Step-by-step explanation:

The slope of a line is the same everywhere, so an equation can be written making use of that fact.

$\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}\\\\\dfrac{y-5}{x-2}=\dfrac{7-5}{6-2}=\dfrac{2}{4}\\\\\boxed{\dfrac{y-5}{x-2}=\dfrac{1}{2}}$

Cross-multiplying gives ...

2(y -5) = x -2

2y -10 = x -2 . . . eliminate parentheses

2y = x +8 . . . . . . . add 10; next, divide by 2

$\boxed{y=\dfrac{1}{2}x+4}$

These equations match choices B and D.

6 0

3 years ago

A gymnasium can hold no more than 650 people.A permanent bleacher in the gymnasium holds 136 people. The event organizers are se

ValentinkaMS [17]

Answer:

20 chairs

Step-by-step explanation:

After 136 people are seated in the bleacher, there can be 514 people seated in chairs. We know that 514 = 25×20 +14, so there can be 20 rows of 25 chairs. We require an equal number of chairs in each row, so there cannot be some rows with 21 chairs, nor can there be a 26th row with 14 chairs.

There can be 20 chairs in each row.

3 0

3 years ago

Read 2 more answers

Which strategy is the most appropriate strategy to solve (x−1)^2=9 ?

Leya [2.2K]

(x-1)^2=9 <=> x-1=3 <=> x=4

3 0

3 years ago

Read 2 more answers

I need help with this plss

baherus [9]

Answer:

2.4 cm.

Step-by-step explanation:

1. Find Midpoint of segment AB which is 10/2 = 5.

2. Find Midpoint of segment BC which is 5.2/2 = 2.6.

3. Subtract both midpoints 5 - 2.6 = 2.4.

That's pretty much it, Hope it helps!

8 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago