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3 years ago

The reciprocal of 4 3/7 is 7/31 true or false?

Mathematics

2 answers:

zloy xaker [14]3 years ago

8 0

False is the answer hope this helps and by the way give the other guy brainliest because on my other account I'm an Ace

docker41 [41]3 years ago

7 0

False, it is 7/28
please give brainliest

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EXPAND THE POLONOMIAL (5a+1/2b)^2 HELPPP ASAP

stepan [7]

Answer:

25 $a^{2}$ +5ab+ $\frac{b^{2} }{4}$

Step-by-step explanation:

$(5a+\frac{1}{2} b)^{2}$

1. First, write out the expression without the square

$(5a+\frac{1}{2} b)(5a+\frac{1}{2} b)$

2. Then use foil to multiply:

5a*5a+5a* $\frac{1}{2}b$ + $\frac{1}{2}b$ *5a+ $\frac{1}{2}b$ * $\frac{1}{2}b$

=25 $a^{2}$ +5ab+ $\frac{b^{2} }{4}$

3. That is your answer.

8 0

3 years ago

Ryan uses 3 bars of clay for a sculpture. The first bar is 1.15 centimeters long. The second bar is 3.92 centimeters long. What

mash [69]

Since we aren't told the length of the 3rd bar, focus solely on the lengths of the first 2 bars. Add together their lengths: 1.15 cm + 3.92 cm = 5.07 cm.

6 0

3 years ago

How do I solve 300x300=

UNO [17]

Answer:

its 90,000

Step-by-step explanation:

just put the zeros together then multiply 3x3

5 0

3 years ago

3m-5p=12 solve for p

umka21 [38]

3m - 5p = 12

3m(-3m) -5p = (-3m)+ 12
-5p = -3m + 12
(-5)/-5p = (-5)/ (-3m + 12)
p = 3/5m + 12/-5

5 0

3 years ago

Read 2 more answers

The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell

emmasim [6.3K]

Answer:

Highest (-3,-3)

Lowest (2,2)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

$\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12$

completing the squares:

$\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2$

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of

$\bf f(x,y)=x^2+y^2$

subject to the constraint

$\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0$

Now we have

$\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)$

Let $\bf \lambda$ be the Lagrange multiplier.

The maximum and minimum must occur at points where

$\bf \nabla f=\lambda\nabla g$

that is,

$\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)$

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

$\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y$

Replacing in the constraint

$\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2$

from this we get

x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0

3 years ago