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2 years ago

In ordered pair (x,y) ,x is called

Mathematics

1 answer:

IceJOKER [234]2 years ago

3 0

Answer:

I think X us called consequent

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in a company, 40% of the workers are women. If 1380 woman work for the company, how many total workers are there?

mrs_skeptik [129]

Answer:

Step-by-step explanation:

The total number of workers is our unknown. If 40% of this unknown number are women and the number of women is 1380, then the equation looks like this:

(remember that the word "of" generally means to multiply)

(also remember that we have to use the decimal form of a percent in an equation)

.40(x) = 1380 then divide to get the number of total workers:

x = 3450

5 0

2 years ago

anthony bought an 8 foot board.He cut off 3/4 of the board to build a shelf,and gave 1/3 of the rest to his brother for an art p

e-lub [12.9K]

Convert your feet to inches.
So 8 times by 12 = 96 inches
if you have 3/4 of the board used.
3/4= .75
if you were to times 96 by .75 = 72 inches.
So you used 72 inches and you have 24 inches left

now do 1/3= .33
times .33 by 24= 8
8 inches is how long you would give to Anthony's brother.

7 0

3 years ago

Read 2 more answers

4. Find the value of the expression below for r=4 and t=2.

Katen [24]

Answer:

(a) 9

Step-by-step explanation:

$\sf t^3 - r + 20 \div r$

substitute r = 4, t = 2

$\sf (2)^3 - 4 + 20 \div 4$

simplify

$\sf (2)^3 - 4 + 5$

cubic of 2 is 8

$\sf 8 - 4 + 5$

simplify

$\sf 9$

5 0

1 year ago

Read 2 more answers

Please Help ASAP!!!!!!

Bumek [7]

Answer: 6

The sides are equal.

Thus

$3x-6=x+2\\2x=8\\x=4$

$JP=x+2=4+2=\boxed{6}$

4 0

2 years ago

Read 2 more answers

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago