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3 years ago

the graph shows rate at which an oven preheats when turned on. what is the unit rate and what does it represent?

Mathematics

2 answers:

zvonat [6]3 years ago

4 0

Answer:

Step-by-step explanation:

i dont know

raketka [301]3 years ago

3 0

Answer:

no lo se está difícil y además nose inglés

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Patricia owned so many posters that she decided to sell 64 of them to her friends. After selling the posters, she still had 176

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Answer:

p - 64 = 176

Step-by-step explanation:

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Given △ABC, sin(A) = A. cos(B) B. sin(B) C. cos(C) D. tan (C)

irinina [24]

the answer is cos(C).

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In any difference 6th grade class the ratio of boys and two girls is 3:2 how many boys and how many girls could there be in one

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3 years ago

Evaluate the expression. 5^2−4⋅6+11 Enter your answer in the box.

mart [117]

Explanation:

This question's it's about the order of operations. P-parenthesis, E-exponents, M-multiply, D-divide, A-add, and S-subtract. It can also go left to right.

Do exponents first.

$5^2=5*5=25$

$25-4*6+11$

Then, multiply left to right.

$4*6=24$

$25-24+11$

And add/subtract left to right.

$25-24=1$

$1+11=12$

$\boxed{=12}$

Hope this helps!

Thanks!

Have a great day!

6 0

3 years ago

Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago