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2 years ago

25x2-8x=12x-4 Can someone please help me on this ?????

Mathematics

2 answers:

kirill [66]2 years ago

7 0

Answer:x=2/5

Step-by-step explanation:

25x2-8x=12x-4

25x2-8x-12x+4=0

25x2-20x+4=0

25x2-10x-10x+4=0

5x(5x-2)-2(5x-2)=0

(5x-2)(5x-2)=0

5x-2=0 and 5x-2=/

5x=2 and 5x=2

X=2/5

Natalka [10]2 years ago

4 0

Answer: X=2.7

Step-by-step explanation:

25*2-8x=12x-4

50-8x = 12x-4

subtract 50 from both sides

-8x = 12x-54

subtract 12x from both sides

-20x= -54

divide both sides by -20

X=2.7

You might be interested in

How to convert 264 yds/1 s to mi/day?

dusya [7]

Answer:

$12,960\ \frac{miles}{day}$

Step-by-step explanation:

we know that

$1\ mile=1,760\ yards$

To convert yards to miles multiply by 1/1,760

$1\ hour=3,600\ seconds$

$1\ day=24\ hours$

$(24)3,600=86,400$

$1\ day=86,400\ seconds$

To convert seconds to day multiply by 1/86,400

In this problem we have

$264\ \frac{yd}{sec} =264\ \frac{(1/1,760)}{(1/86,400)}=264\ (\frac{86,400}{1,760})=12,960\ \frac{miles}{day}$

7 0

2 years ago

How long will it take a $800 investment to be worth $900 if it is continuously compounded at 11% per year?

Kruka [31]

The formula is
A=p e^rt
A future value 900
P present value 800
E constant
R interest rate 0.11
T time?

We need to solve for t
T=[log (A/p)÷log (e)]÷r
T=(log(900÷800)÷log(e))÷0.11
T=1.07 years round your answer to get 1 year

3 0

2 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.