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3 years ago

This is math pls help 3x + 2(5x − 7)

Mathematics

2 answers:

pashok25 [27]3 years ago

7 0

Answer:

13x-14

Step-by-step explanation:

Ira Lisetskai [31]3 years ago

4 0

Answer: 13x - 14

Step-by-step explanation:

3x + 10x - 14

13x -14

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On a coordinate plane, a line goes through (negative 3, 0) and (0, negative 6).

shusha [124]

Answer:

B. (1,-8)

Step-by-step explanation:

Slope of the first one:

(-6-0)/(0--3) = -6/3 = -2

y = -2x - 6

3y + 30 = 6x

y + 10 = 2x

-2x - 6 + 10 = 2x

4 = 4x

x = 1

y = -2(1) - 6

y = -8

8 0

4 years ago

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PLS answer the question below. I will mark brainliest!!

sesenic [268]

Answer:

5 triangles maybe?

Step-by-step explanation:

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3 years ago

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If f(x) = 5x – 2 and g(x) = 2x + 1, find (f- g)(x).

GalinKa [24]

Answer:

la respuesta es A. 3x - 3

7 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).

Here's a comprehensive list of all the critical points we found:

$(\sqrt[4]{13},0,0)$

$(-\sqrt[4]{13},0,0)$

$(0,\sqrt[4]{13},0)$

$(0,-\sqrt[4]{13},0)$

$(0,0,\sqrt[4]{13})$

$(0,0,-\sqrt[4]{13})$

$\left(\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)$

$\left(\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)$

$\left(-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)$

$\left(-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)$

$\left(\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)$

$\left(\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)$

$\left(-\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)$

$\left(-\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)$

$\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

5 0

3 years ago

This 1 seems really complicated

Fofino [41]

The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
________________________________________________________
Given:
________________________________________________________
y = - 4x + 16 ;

4y − x + 4 = 0 ;
________________________________________________________
"Solve the system using substitution" .
________________________________________________________
First, let us simplify the second equation given, to get rid of the "0" ;

→ 4y − x + 4 = 0 ;

Subtract "4" from each side of the equation ;

→ 4y − x + 4 − 4 = 0 − 4 ;

→ 4y − x = -4 ;
________________________________________________________
So, we can now rewrite the two (2) equations in the given system:
________________________________________________________

y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;

4y − x = -4 ; ===> Refer to this as "Equation 2" ;
________________________________________________________
Solve for "x" and "y" ; using "substitution" :
________________________________________________________
We are given, as "Equation 1" ;

→ " y = - 4x + 16 " ;
_______________________________________________________
→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;

to solve for "x" ; as follows:
_______________________________________________________
Note: "Equation 2" :

→ " 4y − x = - 4 " ;
_________________________________________________
Substitute the value for "y" {i.e., the value provided for "y"; in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;

→ as follows:
_________________________________________________

→ " 4 (-4x + 16) − x = -4 " ;
_________________________________________________
Note the "distributive property" of multiplication :
_________________________________________________

a(b + c) = ab + ac ; AND:

a(b − c) = ab <span>− ac .
_________________________________________________
As such:

We have:
</span>
→ " 4 (-4x + 16) − x = - 4 " ;
_________________________________________________
AND:

→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
_________________________________________________
Now, we can write the entire equation:

→ " -16x + 64 − x = - 4 " ;

Note: " - 16x − x = -16x − 1x = -17x " ;

→ " -17x + 64 = - 4 " ; Solve for "x" ;

Subtract "64" from EACH SIDE of the equation:

→ " -17x + 64 − 64 = - 4 − 64 " ;

to get:

→ " -17x = -68 " ;

Divide EACH side of the equation by "-17" ;
to isolate "x" on one side of the equation; & to solve for "x" ;

→ -17x / -17 = -68/ -17 ;

to get:

→ x = 4 ;
______________________________________
Now, Plug this value for "x" ; into "{Equation 1"} ;

which is: " y = -4x + 16" ; to solve for "y".
______________________________________

→ y = -4(4) + 16 ;

= -16 + 16 ;

→ y = 0 .
_________________________________________________________
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
_________________________________________________________
Now, let us check our answers—as directed in this very question itself ;
_________________________________________________________
→ Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten;

→ Let us check;

→ For EACH of these 2 (TWO) equations; do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ;

→ Consider the first equation given in our problem, as originally written in the system of equations:

→ " y = - 4x + 16 " ;

→ Substitute: "4" for "x" and "0" for "y" ; When done, are both sides equal?

→ "0 = ? -4(4) + 16 " ?? ;   → "0 = ? -16 + 16 ?? " ; → Yes! ;

{Actually, that is how we obtained our value for "y" initially.}.

→ Now, let us check the other equation given—as originally written in this very question:

→ " 4y − x + 4 = ?? 0 ??? " ;

→ Let us "plug in" our obtained values into the equation;

{that is: "4" for the "x-value" ; & "0" for the "y-value" ;

→ to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.

→ " 4(0)  − 4 + 4 = ? 0 ?? " ;

→ " 0 − 4 + 4 = ? 0 ?? " ;

→ " - 4 + 4 = ? 0 ?? " ; Yes!
_____________________________________________________
→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
_____________________________________________________
→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
_____________________________________________________
Hope this lenghty explanation is of help! Best wishes!
_____________________________________________________

7 0

3 years ago