All categories

3 years ago

Kim's cookie recipe calls for 1/3 of a cup of sugar. How much sugar would Kim use to make 1/4 of a batch of cookies?

Mathematics

1 answer:

bearhunter [10]3 years ago

3 0

For a quarter of the batch of cookies it requires a quarter of a third
$\frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$

You might be interested in

If log10x = 3, then x =

iren [92.7K]

This is what WolframAlpha says. Look at the image below.

7 0

3 years ago

Read 2 more answers

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago

How do I find the x and the y

mart [117]

Answer:

x = 22

y = 35

Step-by-step explanation:

Because its an isosceles triangle, you know the base angles are equal so:

$3x - 11 = 2x + 11$

Solving:

subtract 2x from both sides

$x-11=11$

add 11 to both sides

$x = 22$

Next, find the angle measures of the two base angles:

$3*22-11\\66-11\\55$

To find y, you know that the interior angles of the whole triangle add up to 180, so you have

$55 + 55 + 2y = 180$

solving,

$110 + 2y = 180$

subtract 110 from both sides

$2y = 70$

divide both sides by 2

$y = 35$

3 0

2 years ago

Read 2 more answers

"The sum of seven-eighths of a<br> number and 20 is 34 at minimum."

Gennadij [26K]

Answer:

Anything greater than or equal to 17.

Step-by-step explanation:

7/8 of 17 is 14.875

6 0

2 years ago

What is the area????

Ad libitum [116K]

Answer:

49.1 cm^2

Step-by-step explanation:

The appropriate area formula is ...

area = (1/2)(side 1)(side 2)(sin(angle between))

= (1/2)(10 cm)(12 cm)·sin(55°) = 60·sin(55°) cm^2

area ≈ 49.1 cm^2

6 0

3 years ago