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1 year ago

A recipe takes 90 minutes

Mathematics

1 answer:

Paha777 [63]1 year ago

4 0

Answer:

Answer

Find out the how many minutes are spent preparing.

To proof

As given

A recipe takes 90 minutes to make.

25 percent of the time is spent preparing the ingredients.

25% is written in the decimal form

= 0.25

Time spent in preparing = 0.25 × 90

= 22.5 minutes

Therefore the 22.5 minutes are spent preparing the ingredients.

Hence proved

Step-by-step explanation:

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How many litters of apple syrup and water did she use in the ratio of 2:3 and made 1800 ml of such fruit juice

sp2606 [1]

Answer:

720 mL ; 1080 ml

Step-by-step explanation:

Givenn:

Apple syrup : water = 2 : 3

Fruit juice volume made using the ratio above = 1800 ml

Sum of ratio = 2 + 3 = 5

Litres of Apple syrup :

(Syrup ratio / total) * total volume of fruit

(2 / 5) * 1800

= 0.4 * 1800

= 720 ml

Litres of water :

(water ratio / total) * total volume of fruit

(3 / 5) * 1800

= 0.6 * 1800

= 1080 ml

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Answer:

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Step-by-step explanation:

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3 years ago

In the derivation of the formula for the volume of a cone, the volume of the cone is calculated to be times the volume of the py

scZoUnD [109]

Answer:

B :it is the ratio of the area of the circle to the area of the square from a cross section.

Step-by-step explanation:

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2 years ago

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For what value of x does 3^4x – 27x-3?

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Step-by-step explanation:

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2 years ago

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This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

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3 years ago