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Brrunno [24]
3 years ago
6

PLEEASEEE HELP ME NOONE WILL HELP ME PLEASE HELP ME

Mathematics
1 answer:
DIA [1.3K]3 years ago
5 0

Answer:

63.97 sheets of paper

Step-by-step explanation:

40-18.25 = 21.75

21.75 divided by 0.35 = 63.97

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) What is the sum?<br> 8+(-3) + (-8) =
uysha [10]

Answer:

-3

Step-by-step explanation:

simplifying, you get 8-3-8, then you get 0-3, then you get -3.

please mark this as brainliest!

5 0
3 years ago
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The ratio of free throws made to free throws tried is 0.75 for a professional basketball player. Last​ year, the player tried 16
andriy [413]
.75 is 75% which is also 3/4 or three-quarters.  They all mean the same thing.  That means he makes 3 out of every 4 shots.
If he shot 16 times he made 12 shots because:
16 X (3/4) = 12
7 0
4 years ago
PLS HELPPP !! BRAINLIEST!!
satela [25.4K]
18cm you multiple it my the figure given
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4 years ago
Verify that the conclusion of Clairaut’s Theorem holds, that is, uxy = uyx, u=tan(2x+3y)
choli [55]

Answer: Hello mate!

Clairaut’s Theorem says that if you have a function f(x,y) that have defined and continuous second partial derivates in (ai, bj) ∈ A

for all the elements in A, the, for all the elements on A you get:

\frac{d^{2}f }{dxdy}(ai,bj) = \frac{d^{2}f }{dydx}(ai,bj)

This says that is the same taking first a partial derivate with respect to x and then a partial derivate with respect to y, that taking first the partial derivate with respect to y and after that the one with respect to x.

Now our function is u(x,y) = tan (2x + 3y), and want to verify the theorem for this, so lets see the partial derivates of u. For the derivates you could use tables, for example, using that:

\frac{d(tan(x))}{dx} = 1/cos(x)^{2} = sec(x)^{2}

\frac{du}{dx}  =  \frac{2}{cos^{2}(2x + 3y)} = 2sec(2x + 3y)^{2}

and now lets derivate this with respect to y.

using that \frac{d(sec(x))}{dx}= sec(x)*tan(x)

\frac{du}{dxdy} = \frac{d(2*sec(2x + 3y)^{2} )}{dy}  = 2*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*3 = 12sec(2x + 3y)^{2}tan(2x + 3y)

Now if we first derivate by y, we get:

\frac{du}{dy}  =  \frac{3}{cos^{2}(2x + 3y)} = 3sec(2x + 3y)^{2}

and now we derivate by x:

\frac{du}{dydx} = \frac{d(3*sec(2x + 3y)^{2} )}{dy}  = 3*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*2 = 12sec(2x + 3y)^{2}tan(2x + 3y)

the mixed partial derivates are equal :)

7 0
3 years ago
A team of three students are working on a language-learning app; they need to develop 300 micro-lessons and 300 micro-tests befo
Olegator [25]

Answer: a) 15 b)

Step-by-step explanation:

Let X be the number of days:

a)

For LESSONS:

Jordan does 10 / day ( 10*X)

Marco 5 / day ( 5*X)

Junyi 5 / day ( 5*X)

For TESTS:

Jordan does 5 / day ( 5*X)

Marco 10 / day ( 10*X)

Junyi 8 / day ( 8*X)

for each they need a total of 300

a) 10X+5X+5X=300 => 20X = 300 => X = 15 days for the lessons

b) 5X+10X+8X = 300 => 23X = 300 => X = 13.04 days for the tests

so they need 15 days to finish both tasks

now if Junyi gets sick we just eliminate his contribution

a) 10X+5X=300 => 15X = 300 => X = 20 days for the lessons

b) 5X+10X = 300 => 15X = 300 => X = 20 days for the tests

so in 20 days they will finish without him

If jordan works 10 hours a day, we just replace him with 10/24

a) 10(10/24)+5X+5X= 300 => X = 29.58 days for the lessons

b) 5(10/24)+10X+8X = 300 => X = 16.51 days for the tests

so at the end to complete both tasks they need 29.58 days

4 0
3 years ago
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