Ask question

Ask question

All categories

3 years ago

8

Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. if x gallons of the 1 percent grade, y gallons of th

e 2 percent grade, and z gallons of the 3 percent grade are mixed to give x + y + z gallons of a 1.5 percent grade, what is x in terms of y and z ?

1 answer:

pshichka [43]3 years ago

6 0

(1x+2y+3z)/(x+y+z)=1.5
x+2y+3z=1.5x+1.5y+1.5z
0.5y+1.5z=0.5x

x=y+3z

You might be interested in

PLEASE HELP ME QUICKLY!!!!!

Gwar [14]

Y=2/5+3; u gotta find two points, count the rise over run and find the y-intercept

3 0

3 years ago

I need the answer for my homework

sergij07 [2.7K]

Answer:

45 for 5 and 34 for 6.

Step-by-step explanation:

You are just subtracting by 11.

7 0

3 years ago

What is the equation of the following direct variation?

nataly862011 [7]

Answer:

y=-2x

Step-by-step explanation:

Because, let us take (-1,2)

2=-2(-1)

2=2

Similarly all the points show the same.

3 0

3 years ago

Read 2 more answers

PLS HELP ME!!! The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of

Pachacha [2.7K]

Hello!

The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³

Data: (Cone)

h (height) = 12 cm

r (radius) = 4 cm (The diameter is 8 being twice the radius)

Adopting: $\pi \approx 3.14$

V (volume) = ?

Solving: (Cone volume)

$V = \dfrac{ \pi *r^2*h}{3}$

$V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}$

$V = 3.14*16*4$

$\boxed{V = 200.96\:cm^3}$

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting: $\pi \approx 3.14$

If: We know that the volume of a sphere is $V = 4* \pi * \dfrac{r^3}{3}$ , but we have a hemisphere, so the formula will be half the volume of the hemisphere $V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}$

Formula: (Volume of the hemisphere)

$V = 2* \pi * \dfrac{r^3}{3}$

Solving:

$V = 2* \pi * \dfrac{r^3}{3}$

$V = 2*3.14 * \dfrac{4^3}{3}$

$V = 2*3.14 * \dfrac{64}{3}$

$V = \dfrac{401.92}{3}$

$\boxed{ V_{hemisphere} \approx 133.97\:cm^3}$

What is the approximate volume of this figure?

Now, to find the total volume of the figure, add the values: (cone volume + hemisphere volume)

Volume of the figure = cone volume + hemisphere volume

Volume of the figure = 200.96 cm³ + 133.97 cm³

$\boxed{\boxed{\boxed{V = 334.93\:cm^3 \to Volume\:of\:the\:figure \approx 335\:cm^3 }}}\end{array}}\qquad\quad\checkmark$

Answer:

The volume of the figure is approximately 335 cm³

_______________________

I Hope this helps, greetings ... Dexteright02! =)

8 0

3 years ago

Read 2 more answers

Use log 2=a, log 3=b to evaluate log 216

Anuta_ua [19.1K]

Answer:

answer = 3(a +b)

Step-by-step explanation:

$log(2) = a \: log(3) = b \: \: \\ log(216) = log(2 \times 2 \times 2 \times 3 \times 3 \times 3) \\ = log( {2}^{3} \times {3}^{3} ) \\ = log(2^{3} ) + log( {3}^{3} ) \\ = 3 log(2) + 3 log(3) \\ = 3(a) + 3(b) \\ = 3(a + b) \\ \\ log(216) = 3(a + b)$

8 0

2 years ago