All categories

3 years ago

Pierre uses 30 grams of chocolate to make

Mathematics

1 answer:

Travka [436]3 years ago

3 0

Pierre can make 3 liters of chocolate sauce from 450 grams of chocolate.

Step-by-step explanation:

Given,

Chocolate used to make 200 ml of chocolate sauce = 30 grams

Ratio of chocolate to milliliters of chocolate sauce = 30:200

Let,

x be the milliliters of sauce made from 450 grams of chocolate

Ratio of chocolate to ml of sauce = x:450

Using proportion;

Ratio of chocolate to ml of sauce:: ratio of chocolate to ml of sauce

$30:200::450:x$

Product of extreme = Product of mean

$30*x=200*450\\30x=90000$

Dividing both sides by 30

$\frac{30x}{30}=\frac{90000}{30}\\x=3000$

450 grams of chocolate will make 3000 milliliters of chocolate sauce.

We will convert this in liters.

1 liter = 1000 ml

3000 ml = $\frac{3000}{1000}$ = 3 liters

Pierre can make 3 liters of chocolate sauce from 450 grams of chocolate.

Keywords: ratio, proportion

Learn more about proportion at:

#LearnwithBrainly

You might be interested in

Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction p

kirza4 [7]

The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

<h3>How to convert sine of an angle to some angle of cosine?</h3>

We can use the fact that:

$\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)$

to convert the sine to cosine.

<h3>Which trigonometric functions are positive in which quadrant?</h3>

In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
In second quadrant(π/2 < θ < π), only sin and cosec are positive.
In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

$y= 3\cos(2(x + \pi/2)) - 2$

The options are:

$y= 3\sin(2(x + \pi/4)) - 2$
$y= -3\sin(2(x + \pi/4)) - 2$
$y= 3\cos(2(x + \pi/4)) - 2$
$y= -3\cos(2(x + \pi/2)) - 2$

Checking all the options one by one:

Option 1: $y= 3\sin(2(x + \pi/4)) - 2$

$y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2$

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

Option 2: $y= -3\sin(2(x + \pi/4)) - 2$

This option if would be true, then from option 1 and this option, we'd get:
$-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0$

which isn't true for all values of x.

Thus, this option is not same as the given function.

Option 3: $y= 3\cos(2(x + \pi/4)) - 2$

The given function is $y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2$

This option's function simplifies as:

$y= 3\cos(2(x + \pi/4)) - 2 = 3\cos(2x + \pi/2) -2 = -3\sin(2x) - 2$

Thus, this option isn't true since $\sin(2x) \neq \cos(2x)$ always (they are equal for some values of x but not for all).

Option 4: $y= -3\cos(2(x + \pi/2)) - 2$

The given function simplifies to: $y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2$

The given option simplifies to:

$y= -3\cos(2(x + \pi/2)) - 2 = -3\cos(2x + \pi ) -2\\y = 3\cos(2x) -2$

Thus, this function is not same as the given function.

Thus, the function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

Learn more about sine to cosine conversion here:

brainly.com/question/1421592

4 0

2 years ago

Read 2 more answers

Which expression simplifies...

Zina [86]

Answer:

see below

Step-by-step explanation:

If suffices to look at the last term. It must simplify to 4/(3w²), so the exponents of z must match in numerator and denominator, and the ratio of constants must reduce to 4/3.

The exponents of z have the right values in choices B and D, but the ratio of constants only matches in choice D. Looking at the rest of choice D, we see that we can factor 6w²z² from both numerator and denominator, then factor 2 from the remaining numerator to make it match the desired form.

$\dfrac{(12w^2z^4-24w^2z^2)}{18w^4z^2}=\dfrac{6w^2z^2}{6w^2z^2}\cdot\dfrac{2z^2-4}{3w^2}=\dfrac{2(z^2-2)}{3w^2}$

3 0

3 years ago

If line q bisects EG at point T, and if ET= 1/3 x and TG= x-2, then find EG

kvasek [131]

Answer:

EG = 2 units

Step-by-step explanation:

Given that line q bisects EG at T , then

ET = TG ( substitute values )

$\frac{1}{3}$ x = x - 2 ( multiply through by 3 to clear the fraction )

x = 3x - 6 ( subtract x from both sides )

0 = 2x - 6 ( add 6 to both sides )

6 = 2x ( divide both sides by 2 )

3 = x

Then

ET = $\frac{1}{3}$ x = $\frac{1}{3}$ × 3 = 1

TG = x - 2 = 3 - 2 = 1

Thus

EG = ET + TG = 1 + 1 = 2 units

8 0

2 years ago

Read 2 more answers

Item 20 A rectangular school banner has a length of 54 inches and a width of 36 inches. A sign is made that is similar to the sc

siniylev [52]

Answer:

The ratio of the area of the school banner to the area of the sign is <u>1944 cubic inches : 192.61 cubic inches.</u>

Step-by-step explanation:

Given:

A rectangular school banner has a length of 54 inches and a width of 36 inches. A sign is made that is similar to the school banner and has a length of 17 inches.

Now, to find the ratio of the area of the school banner to the area of the sign.

Dimensions of school banner :

Length = 54 inches.

Width = 36 inches.

Dimension of school sign:

Length = 17 inches.

So, to we find the width of sign by using cross multiplication method:

Let the width be $x.$