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2 years ago

PLZZ HELP I NEED HELP

Mathematics

1 answer:

Mkey [24]2 years ago

3 0

Step-by-step explanation:

1) let the number=x

six times a number=6x

Condition:

6x+4=22

2) eleven times a number=11x

Condition:

11x-5=50

3) 9 times a number=9x

Condition:

9x-7=-16

Note:if you need to ask any question please let me know.

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Find the distance between these two lines: 4x+3y=7 and 4x+3y=27

Alex73 [517]

There is no distance since they are the same exact number.

4 0

3 years ago

Mark and Julio are selling flower bulbs for a school fundraiser. Customers can buy bags of windflower bulbs and packages of croc

ikadub [295]

Answer:

Bag of windflower bulbs costs $8.50

Package of crocus bulbs costs $17.60

Step-by-step explanation:

Let $x be the price of one bag of windflower bulbs and $y be the price of one package of crocus bulbs.

1. Mark sold 2 bags of windflower bulbs for $2x and 5 packages of crocus bulbs for $5y. In total he earned $(2x+5y) that is $105. So,

2x+5y=105

2. Julio sold 9 bags of windflower bulbs for $9x and 5 packages of crocus bulbs for $5y. In total he earned $(9x+5y) that is $164.50. So,

9x+5y=164.50

3. You get the system of two equations:

$\left\{\begin{array}{l}2x+5y=105\\ \\9x+5y=164.50\end{array}\right.$

From the first equation

$5y=105-2x$

Substitute it into the second equation:

9x+105-2x=164.50

7x=164.50-105

7x=59.5

x=$8.50

So,

5y=105-2·8.5

5y=105-17

5y=88

y=$17.60

5 0

3 years ago

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$