2 years ago

Solve the following equation. Show your work. 7x + 14 = 7(x+2)

Mathematics

2 answers:

Mama L [17]2 years ago

8 0

7(x+2)= 7x + 14
7x + 14 = 7x + 14

Solnce55 [7]2 years ago

6 0

The equations are the same, so both the lines are the same and thus have infinite answers

You might be interested in

F(x+k) what is the value of k

guapka [62]

Answer:

k = -3

Hope this helps!

6 0

2 years ago

Read 2 more answers

-7d=56<br><img src="https://tex.z-dn.net/?f=%20-%207%20%3D%2056" id="TexFormula1" title=" - 7 = 56" alt=" - 7 = 56" align="absmi

PtichkaEL [24]

The only thing you need to do is divide 56 by -7. You know the d must be a negative number because negative x negative= positive. So 56 / -7 is -8.

7 0

2 years ago

Ross needs to have a procedure done. Because he is over 65, he is on medicare. Their plan says that according to their plan thei

artcher [175]

300=(1/6*7/8)x
300=7/48x
300*48=7x
14400=7x
2057=x

So the answer is B

Hope this helps :)

4 0

3 years ago

Read 2 more answers

This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex

noname [10]

$L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)$

$L_x=10+10\lambda x=0\implies1+\lambda x=0$

$L_y=10+10\lambda y=0\implies1+\lambda y=0$

$L_z=2+4\lambda z=0\implies1+2\lambda z=0$

$L_\lambda=5x^2+5y^2+2z^2-42=0$

$\begin{cases}L_x=0\\L_y=0\\L_z=0\end{cases}\implies1+\lambda x=1+\lambda y=1+2\lambda z=0\implies x=y=2z$

$5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1$

$z^2=1\implies z=\pm1\implies x=y=\pm2$

There are two critical points, at which we have

$f\left(2,2,1\right)=42\text{ (a maximum value)}$

$f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}$

3 0

3 years ago

Please answer thank you!

vivado [14]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers