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3 years ago

Find the distance between the points (-10, -4) and (-4,-4).

Mathematics

1 answer:

Rudiy273 years ago

3 0

Answer:

im here for the points

Step-by-step explanation:

You might be interested in

The second of two numbers is 3 less than twice the first. Their sum is 36. Find the numbers.

o-na [289]

Answer:

13 and 23

Step-by-step explanation:

create an equation to solve:

(2x-3) + x = 36

3x - 3 = 36 combine like terms

+3 +3 add 3 to both sides

3x = 39

3 3 divide both sides by 3

x = 13

13 is the first number

the 2nd number is 3 let than twice the first so we create and solve a new equation:

2x - 3 = ?

2(13) = ? substitute 13 from the

smaller number to solve

23 = the 2nd number

3 0

4 years ago

What is 1035 divide 12

VikaD [51]

Answer:

86.25

If you go on desmos scientific calculator it makes it super easy

4 0

3 years ago

Read 2 more answers

With the air conditioner on, the electric bill was $303. With the air conditioner off, the bill was $194. What was the differenc

professor190 [17]

Answer:

$109 is the difference between both the bills

Step-by-step explanation:

Difference =$(303-194)

=$109

Hope this helps

6 0

3 years ago

Read 2 more answers

X+7+=-14 ASAP! PLease!

umka21 [38]

Hey there!

Answer = -21

Explanation:

Let's solve your equation step-by-step.

x + 7 = -14

Subtract seven from both sides:

x + 7 - 7 = -14 - 7

x = -21

So therefore, that is how you found your answer: x = -21

8 0

4 years ago

Read 2 more answers

Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.

irina1246 [14]

Lagrange multipliers:

$L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)$

$L_x=y^2z^2+\lambda=0$
$L_y=2xyz^2+\lambda=0$
$L_z=2xy^2z+\lambda=0$
$L_\lambda=x+y+z-5=0$

$\lambda=-y^2z^2=-2xyz^2=-2xy^2z$

$-y^2z^2=-2xyz^2\implies y=2x$ (if $y,z\neq0$ )

$-y^2z^2=-2xy^2z\implies z=2x$ (if $y,z\neq0$ )

$-2xyz^2=-2xy^2z\implies z=y$ (if $x,y,z\neq0$ )

In the first octant, we assume $x,y,z>0$ , so we can ignore the caveats above. Now,

$x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2$

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of $f(1,2,2)=16$ .

We also need to check the boundary of the region, i.e. the intersection of $x+y+z=5$ with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force $f(x,y,z)=0$ , so the point we found is the only extremum.

4 0

4 years ago