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SOVA2 [1]
3 years ago
5

Need help on 2 plz I can't understand it

Mathematics
1 answer:
Sphinxa [80]3 years ago
4 0
-3 1/2 
-7/8(change) * 4(days)= -3 1/2 or -3.5
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Find m∠4 if m∠1 = 100°
podryga [215]

Answer:

m4{=100 is the answer,,,,,,

5 0
3 years ago
Three times the perimeter of a triangle is the same as 75 decreased by twice the perimeter. What is the perimeter of the triangl
kupik [55]
Solution
let p be the perimeter of the triangle then
3p=75-2p   collect the like terms
3p+2p=75
5p=75  divide both side by 5
5p/5=75/5⇒p=15.
therefore the perimeter of triangle is 15 units  
6 0
4 years ago
Shannon is putting a fence around the garden, except where there is a gate that is 3 feet wide. One foot of the fence costs $43.
Inessa [10]

Answer:

The expression would be 43x + 128

Step-by-step explanation:

43 dollars is multiplied by however many feet is used.  Then, the exception(the gate) is added to the charge that is 43x

8 0
3 years ago
Consider the function f(x, y) = x2 + xy + y2 defined on the unit disc, namely, d = {(x, y)| x2 + y2 ≤ 1}. use the method of lagr
Pani-rosa [81]
First we note that the partial derivatives vanish simultaneously at one point:

\begin{cases}f_x=2x+y=0\\f_y=x+2y=0\end{cases}\implies(x,y)=(0,0)

so there is one critical point within the region D.

The Lagrangian is

L(x,y,\lambda)=x^2+xy+y^2+\lambda(x^2+y^2-1)

and has partial derivatives

L_x=2x+y+2\lambda x
L_y=x+2y+2\lambda y
L_\lambda=x^2+y^2-1

Set each partial derivative to 0 to find the possible critical points within the disk D. Then we notice that

yL_x=2xy+y^2+2\lambda xy=0
xL_y=x^2+2xy+2\lambda xy=0
L_\lambda=0\implies x^2+y^2=1

\implies xL_y-yL_x=x^2-y^2=0

Since x^2+y^2=1, we have

x^2-y^2=x^2+y^2-2y^2=0\implies1=2y^2\implies y^2=\dfrac12\implies y=\pm\dfrac1{\sqrt2}

And since x^2-y^2=0, or x^2=y^2, we also have

x=\pm\dfrac1{\sqrt2}

So we have four possible additional critical points to consider:

f(0,0)=0
f\left(-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac32
f\left(-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac12
f\left(\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac12
f\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac32

It should be clear enough which of these correspond to the absolute extrema of f over D.
8 0
3 years ago
Simplify the expression3х + 8 + 13x =
leva [86]

Answer:

8(2x+1)

Step-by-step explanation:

First we add

16x + 8

Next take GCF

8(2x+1)

6 0
3 years ago
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