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1 year ago

How many 1/2 cup servings are in 9 cups of cereal?

Mathematics

1 answer:

Arlecino [84]1 year ago

5 0

Answer:

Step-by-step explanation:

9 divided by 1/2 is the same as 9 multiplied by 2, which equals 18

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The roof of this garden shed has a gradient of 0.35. find the height of the shed

Ilya [14]

The height of the shed is 2.325 m

The complete question is

The roof of a lean-to-garden shed has a gradient of 0.35.find the height....they have given the base as 1.5m and a shorter height 1.8m its like a house

<h3>What is a right Triangle ?</h3>

In a Triangle when one angle is 90 degree , the triangle is called Right Triangle .

Structure of the house is a right triangle with base = 1.5m

(height of that triangle)/1.5 = 35/100

height of that triangle = 1.5*0.35 = 0.5250

Ans: greater height of the lean-to = 1.8 + 0.5250 = 2.325 meters.

The height of the shed is 2.325 m.

To know more about Right Triangle

brainly.com/question/6322314

#SPJ1

6 0

2 years ago

Show work <br> Need fast!!! <br><br> 4(1/3+ 1/3+ 1/3)

aalyn [17]

Answer:

Step-by-step explanation:

calculate the sum 4*1 any expression is multiplied by 1 remains the same solution is 4

7 0

3 years ago

What the volume is of this solid

mrs_skeptik [129]

Answer:

336.

Step-by-step explanation:

8 x 7 x 6 = 336.

Feel free to let me know if you need more help! :)

5 0

2 years ago

Harman [31]

Answer: an improper fraction

Step-by-step explanation: An improper fraction is a fraction that is larger than 1 7/3 is larger than one because 7, the numerator is larger than the denominator so 7/3 is an improper fraction

3 0

1 year ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

1 year ago