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3 years ago

A boy is 5 years old and his sister is three times as old as he is.

Mathematics

2 answers:

Gemiola [76]3 years ago

7 0

The sister is 15 and the brother is 5 so add 5 years on top of 15 and you get the sisters present age which is 20

hichkok12 [17]3 years ago

3 0

His sister will be 20, right!

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gayaneshka [121]

It would be 60 mm ( i think )

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3 years ago

3) Which property is used in the following: 4 x (5 + 7) = 4 x 5 + 4x7

earnstyle [38]

Answer:

3) D

4) B

5) A

6) 2 (x + 7)

Step-by-step explanation:

D. Distributive Property

The distributive property of multiplication over addition can be used when you multiply a number by a sum.

B. 2 + 8 = 8 + 2

The word "commutative" comes from "commute" or "move around", so the Commutative Property is the one that refers to moving stuff around.

A. 4 x 1 = 4

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3 ( x + 3) - X + 5

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3 years ago

-1•f(-9)+7•g(6)= A.32 B.34 C.36 D.38

pishuonlain [190]

Answer:

Step-by-step explanation:

-1•f(-9)+7•g(6)

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g(6) = 6

-1* 4+7*6

-4 +42

4 0

3 years ago

A father wants to buy a total of five milk drinks for his son and spend $7.95. Eggnog costs $1.55 and a peanut punch costs $1.65

tresset_1 [31]

Answer:

Step-by-step explanation:

$ 1.55 + $1.65 = $3.11

the budget is $7.95

$7.95 ÷ $3.11 = 2.5 but since you can't get half a milk it's 2

6 0

3 years ago

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:

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

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

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

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 $e^c = C$

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

Now from the question we are told that

$y(0) = y_o$

Hence

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

=> $y_o = C$

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

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

$x(0) = x_o$

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

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

divide both side by $(K * x_o)$

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

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

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

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

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

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

5 0

3 years ago