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3 years ago

Marina’s bread recipe calls for 2/3 of a cup of flour. Which amount of flour represents the same amount?

Mathematics

1 answer:

Montano1993 [528]3 years ago

5 0

Answer:

4/6

Step-by-step explanation:

this is going to be your answer because 2/6 is 1/3 cup and then 4/6 is 2/3 of a cup.

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Write the ratio: <br> 450<br> to <br> 200<br> as a fraction in lowest terms.

xz_007 [3.2K]

Answer:

The ratio 450 : 200 can be reduced to lowest terms by dividing both terms by the GCF = 50 :

450 : 200 = 9 : 4

Therefore:

450 : 200 = 9 : 4

basically 9/4

Step-by-step explanation:

6 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

A pyramid has a base area of 124 in² and a height of 25.2 in. What is the volume of the pyramid?

xeze [42]

Answer:

1041.6

Step-by-step explanation:

The volume of a pyramid is V=b x h x 1/3 . b is the area of the base and h is the height. V=124 x 25.2 x 1/3 This can simplify to V=3124.8 x 1/3 . You could also just divide by 3. You will get 1041.6

8 0

3 years ago

If an open box has a square base and a volume of 115 in.3 and is constructed from a tin sheet, find the dimensions of the box, a

Karolina [17]

Let h = height of the box,
x = side length of the base.

Volume of the box is $V=x^{2} h = 115$ .
So $h = \frac{115}{ x^{2} }$

Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
$S = 2( x^{2} + hx + hx) \\ = 2 x^{2} + 4hx \\ = 2 x^{2} + 4( \frac{115}{ x^{2} } )x$
$= 2 x^{2} + \frac{460}{x}$
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.

$S' = 4x- \frac{460}{ x^{2} } = 0$
$4x = \frac{460}{ x^{2} } \\ 4x^{3} = 460 \\ x^{3} = 115 \\ x = \sqrt[3]{115} = 4.86$
Then $h= \frac{115}{4.86^{2}} = 4.87$
So the box is 4.86 in. wide and 4.87 in. high.

5 0

3 years ago

Read 2 more answers

The magnitude of vector a = (5,m) is 13 and the magnitude of vector b = (n, 24) is 25. What are m and n

Ulleksa [173]

Answer:

$m=12\,,n=7$