Help me with this I don't get it help4help?

Mathematics

1 answer:

Aleks [24]3 years ago

6 0

1) Divide 3/4 by 1/2. Remember how to divide fractions. Switch the problem into a multiplication problem.
3/4 x 2/1 = 3/2 (so you could only have 1 and 1/2 servings with 1/2 pound of chicken in each serving. 

2) Set up a little algebra problem. Let x = the multiplier of the original 3/4 pounds of chicken needed to make two full servings with 1/2 pound of chicken in each. 
(3/4)x = 2(1/2) (Found this online V(0-0)v)

You might be interested in

A water tank holds 648 gallons but is leaking at a rate of 4 gallons per week. A second water tank holds 864 gallons but is leak

Paraphin [41]

Answer:

64 weeks, I hope it helps :)

5 0

3 years ago

8th GRADE MATH

s2008m [1.1K]

The answer for the question is a

6 0

2 years ago

Read 2 more answers

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 $y$

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