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4 years ago

Telling of a classroom is 32 feet. The width is 75% of the lengt. Find the area of the classroom in square feet

Mathematics

1 answer:

Evgesh-ka [11]4 years ago

7 0

The area of the classroom is 768 square feet

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Some one please help me Thx

seraphim [82]

Answer:

3/5

Step-by-step explanation:

3 of the 5 are shaded or indicated

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3 years ago

Read 2 more answers

A decreased by 20% and then increased by 20% express using algebra<br> asapppppp

Ket [755]

Answer:

96%

Step-by-step explanation:

assume the whole number is x

the first thing is decreasing by 20 %

x - 20%x = 80% x

then now the 80%x is the whole number

now It's the turn of increasing

80%x + 20% * (80% x) = 80%x + 16%x = 96% x

I hope this is hlepful to you !!!!

Feel free to ask me in the comments

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3 years ago

What is the factor of 40,24,16

nekit [7.7K]

There are many factors of 40, 24, and 16

All three numbers are divisible by 2, 4, and 8

Hope this helps

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3 years ago

Micah is buying items for a birthday party. If he wants to have the same amount of each item, what is the least number of packag

Bezzdna [24]

2 packages , that’s the answer .

7 0

3 years ago

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Rewrite the statement in mathematical notation. (Let y be the distance from the top of the ladder to the floor, x be the distanc

In-s [12.5K]

Answer:

$\frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}$

Step-by-step explanation:

Let L be the length of the ladder,

Given,

x = the distance from the base of the ladder to the wall, and t be time.

y = distance from the base of the ladder to the wall,

So, by the Pythagoras theorem,

$L^2 = y^2 + x^2$

$\implies L = \sqrt{y^2 + x^2}$ ,

Differentiating with respect to time (t),

$\frac{dL}{dt}=\frac{d}{dt}(\sqrt{x^2 + y^2})$

$=\frac{1}{2\sqrt{x^2 + y^2}}\frac{d}{dt}(x^2 + y^2)$

$=\frac{1}{2\sqrt{x^2 + y^2}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})$

$=\frac{1}{\sqrt{x^2 +y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Here,

$\frac{dy}{dt}=-6\text{ ft per sec}$

Also, $\frac{dL}{dt} = 0$ ( Ladder length = constant ),

$\implies \frac{1}{\sqrt{x^2 +y^2}}(x(-6)+y\frac{dy}{dt})=0$

$-6x + y\frac{dy}{dt}=0$

$y\frac{dy}{dt}=6x$

$\implies \frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}$

Which is the required notation.

8 0

3 years ago