How can you use unit rates in multiplication to solve for missing measures and equivalent ratio problems?

1 answer:

4 0

By viewing equivalent ratios and rates as deriving from, and extending, pairs of rows (or columns) in the multiplication table and by analyzing simple drawings that indicate the relative size of quantities.

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Twice a number divided by 6 is 42

lakkis [162]

X = a number

2x/6 = 42

Isolate the x, first multiply 6 to both sides

2x/6(6) = 42(6)

2x = 42(6)

2x = 252

Isolate the x, divide 2 from both sides

2x/2 = 252/2

x = 252/2

x = 126

126 is your answer

hope this helps

7 0

3 years ago

Read 2 more answers

There are a total of 64 students in a drama club and a yearbook club. The drama club has 10 more students than the yearbook club

Lynna [10]

Drama+yearbook=64
drama=10 more than yearbook
drama=10+yearbook

the equations are
x+y=64
x=10+y

solve
subsitue 10+y for x
10+y+y=64
10+2y=64
minus 10 both sides
2y=54
divide both sides by 2
y=27

subsitute back
x=10+y
x=10+27
x=37

37 people in yearbook club
27 in drama club

5 0

3 years ago

Solve the equation x^3 y^'" + 5x^2 y" + 7xy' + 8y = 0

seraphim [82]

Make the substitution $t=\ln x$ , then compute the derivatives of $y$ with respect to $t$ via the chain rule.

First derivative

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}$

Second derivative

Let $f(t)=\frac{\mathrm dy}{\mathrm dt}$ .

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac fx\right]=\dfrac{x\frac{\mathrm df}{\mathrm dx}-f}{x^2}$

$\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dt^2}$

$\implies\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}\right)$

Third derivative

Let $g(t)=\frac{\mathrm df}{\mathrm dt}=\frac{\mathrm d^2y}{\mathrm dt^2}$ .

$\dfrac{\mathrm d^3y}{\mathrm dx^3}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{g-f}{x^2}\right]=\dfrac{x^2\left(\frac{\mathrm dg}{\mathrm dx}-\frac{\mathrm df}{\mathrm dx}\right)-2x(g-f)}{x^4}$

$\dfrac{\mathrm dg}{\mathrm dx}=\dfrac{\mathrm dg}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^3y}{\mathrm dt^3}$

$\implies\dfrac{\mathrm d^3y}{\mathrm dx^3}=\dfrac{x^2\left(\frac1x\frac{\mathrm dg}{\mathrm dt}-\frac1x\frac{\mathrm df}{\mathrm dt}\right)-2x(g-f)}{x^4}=\dfrac1{x^3}\left(\dfrac{\mathrm d^3y}{\mathrm dt^3}-3\dfrac{\mathrm d^2y}{\mathrm dt^2}+2\dfrac{\mathrm dy}{\mathrm dt}\right)$

Substituting $y(t)$ and its derivatives into the ODE gives a new one that is linear in $t$ :

$\left(\dfrac{\mathrm d^3y}{\mathrm dt^3}-3\dfrac{\mathrm d^2y}{\mathrm dt^2}+2\dfrac{\mathrm dy}{\mathrm dt}\right)+5\left(\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}\right)+7\dfrac{\mathrm dy}{\mathrm dt}+8y=0$