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LUCKY_DIMON [66]

2 years ago

11

1whole and 1 half plus 3 and 2 thirds

2 answers:

almond37 [142]2 years ago

7 0

Add a 3ed to the pie

guapka [62]2 years ago

5 0

The correct answer would be 5/16 because 1 plus 1/2 equals 1 and a half and 3 plus two-thirds equals 3 and 2 thirds. Then you find the LCM which is 16, multiply the numerators by the factors and you get 5/16.

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What is the length of BC in the right triangle below?

sleet_krkn [62]

Answer:

F. 82

Step-by-step explanation:

To calculate the length of BC, Pythagoras theorem will be used.

It says, Hypotenuse^2 = Perpendicular^2 + Base^2

In our case, Perpendicular = AB = 18 and Base = AC = 80

therefore,

Hypotenuse^2 = 18^2 + 80^2 = 6724

Hypotenuse = BC = $\sqrt{6724}$ = 82

6 0

2 years ago

Hellooo i'm an 8th grader in an algebra 1 class so i will be clueless but my problem is

Shalnov [3]

Add 3x to both sides of the equation which means y=13+3x

3 0

2 years ago

Read 2 more answers

Hello if yall can help me please...

nydimaria [60]

B. 3

at 4 weeks it is at 12 cm. 4x3 is 12

3 0

2 years ago

Read 2 more answers

The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli i

harina [27]

Answer:

a

$y(t) = y_o e^{\beta t}$

b

$x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

c

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

Step-by-step explanation:

From the question we are told that

$\frac{dy}{y} = -\beta dt$

Now integrating both sides

$ln y = \beta t + c$

Now taking the exponent of both sides

$y(t) = e^{\beta t + c}$

=> $y(t) = e^{\beta t} e^c$

Let $e^c = C$

So

$y(t) = C e^{\beta t}$

Now from the question we are told that

$y(0) = y_o$

Hence

$y(0) = y_o = Ce^{\beta * 0}$

=> $y_o = C$

So

$y(t) = y_o e^{\beta t}$

From the question we are told that

$\frac{dx}{dt} = -\alpha xy$

substituting for y

$\frac{dx}{dt} = - \alpha x(y_o e^{-\beta t })$

=> $\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt$

Now integrating both sides

$lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c$

Now taking the exponent of both sides

$x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}$

=> $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c$

Let $e^c = A$

=> $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now from the question we are told that

$x(0) = x_o$

So

$x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

=> $x_o = K e^{\frac {\alpha y_o }{\beta } }$

divide both side by $(K * x_o)$

=> $K = x_o e^{\frac {\alpha y_o }{\beta } }$

So

$x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

=> $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=> $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

Generally as t tends to infinity , $e^{- \beta t}$ tends to zero

so

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

5 0

3 years ago

Translate the following phrase into an algebraic expression. Use the variable b to represent the unknown quantity.

otez555 [7]

Answer:

s/5=b

Step-by-step explanation:

4 0

2 years ago