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3 years ago

In which quadrant is the solution to this system of equations? 2x + 5y = 1 3x - 4y = 13

Mathematics

1 answer:

Reptile [31]3 years ago

4 0

Answer: Quadrant IV

Step-by-step explanation: Pic attached.

I found the x and y using substitution (isolating one variable and plugging it into the other). There was probably an easier way but I couldn't think of it at that moment.

From there, you have the point ( 133/36 , -23/18), or a (+, -).

Find Which quadrant has positive x and negative y.

In this case, this would be Quadrant IV or 4.

Hope this helps! Sorry if too complicated lol.

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Booneville Middle School spends $22 for every workbook it buys. At most how many

Kipish [7]

Answer:

Step-by-step explanation:

$44 / 22 = 2

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3 years ago

you are going to buy doughnuts for your coworkers. you buy 2 boxes of glazed doughnuts for $3.99 each, 3 boxes of chocolate-cove

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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

olga nikolaevna [1]

$f(x_1,\ldots,x_n)=x_1+\cdots+x_n=\displaystyle\sum_{i=1}^nx_i$

${x_1}^2+\cdots+{x_n}^2=\displaystyle\sum_{i=1}^n{x_i}^2=4$

The Lagrangian is

$L(x_1,\ldots,x_n,\lambda)=\displaystyle\sum_{i=1}^nx_i+\lambda\left(\sum_{i=1}^n{x_i}^2-4\right)$

with partial derivatives (all set equal to 0)

$L_{x_i}=1+2\lambda x_i=0\implies x_i=-\dfrac1{2\lambda}$

for $1\le i\le n$ , and

$L_\lambda=\displaystyle\sum_{i=1}^n{x_i}^2-4=0$

Substituting each $x_i$ into the second sum gives

$\displaystyle\sum_{i=1}^n\left(-\frac1{2\lambda}\right)^2=4\implies\dfrac n{4\lambda^2}=4\implies\lambda=\pm\frac{\sqrt n}4$

Then we get two critical points,

$x_i=-\dfrac1{2\frac{\sqrt n}4}=-\dfrac2{\sqrt n}$

$x_i=-\dfrac1{2\left(-\frac{\sqrt n}4\right)}=\dfrac2{\sqrt n}$

At these points we get a value of $f(x_1,\cdots,x_n)=\pm2\sqrt n$ , i.e. a maximum value of $2\sqrt n$ and a minimum value of $-2\sqrt n$ .

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3 years ago

How much should 5x³+3x²-2x+1 be increased to get 6x² +7

Mnenie [13.5K]

Answer:

-5x³ + 3x³ + 2x + 6

Step-by-step explanation:

if that is truly the complete problem description, then the only possible answer I can think of, that makes any sense :

what do I need to add to 5x³+3x²-2x+1 to get 6x²+7 ?

the answer is

6x²+7 - (5x³+3x²-2x+1) = 6x²+7-5x³-3x²+2x-1 =

= -5x³ + 3x³ + 2x + 6

5 0

3 years ago

hapis has four buckets. The buckets hold 4700ml,4.8l,3880ml,and 4.71 l hapis wants to to use the bucket that holds the closest t

Mariana [72]

Answer:

Bucket 2

Step-by-step explanation:

Bucket 1 = 4700ml

Bucket 2 = 4.8l

Bucket 3 = 3880ml

Bucket 4 = 4.71 l

Convert holding capacity of bucket 1 and bucket 3 into liters

1 liter = 1000 milliliter

Bucket 1 = 4700ml

4700ml = 4.7l

Bucket 3 = 3880ml

3880ml = 3.88l

Therefore,

Bucket 1 = 4.7l

Bucket 2 = 4.8l

Bucket 3 = 3.88l

Bucket 4 = 4.71 l

The bucket closest to 5liters is bucket 2 with a holding capacity of 4.8 liters

4 0

3 years ago