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Valentin [98]
1 year ago
15

Which numbers I have to put in the boxes

Mathematics
1 answer:
djyliett [7]1 year ago
5 0
Where’s the problem
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Which set of numbers could represent the lengths of the sides of a right triangle?
vitfil [10]

in this problem what you are really looking for is which of these sets is a pathagorean triple. That means it will solve the pathagorean theorem. (a sqaured + b squared = c squared) c is always going to be the largest number or the hypotenuse. if you plug all the number sets into the theorem, only one works and that is 7, 24, 25 which is your answer.

4 0
3 years ago
Rewrite the following polynomial so the exponents decrease from left to right 22x4 – 3x8 + 12Rewrite the following polynomial so
Cloud [144]
-3x^8+22x^4+12
5 0
4 years ago
What is √101.6064 in to 1 decimal place and in standard form?​<br>NEED ASAP
Dimas [21]

Note that

10² = 100

0.08² = 0.0064

and

101.6064 = 100 + 1.6 + 0.0064

It also happens that

1.6 = 2 • 10 • 0.08

which means

101.6064 = 100 + 1.6 + 0.0064

101.6064 = 10² + 2 • 10 • 0.08 + 0.08²

101.6064 = (10 + 0.08)²

101.6064 = 10.08²

which means

√101.6064 = 10.08

which, to one decimal place, is approximately 10.1.

3 0
3 years ago
Find the area of the polygon ​
katrin [286]

Answer:

just add the area of the triangle(formula 1/2bh) and the area of the square(formula bh)

so 1/2(6*height)+8*7

Step-by-step explanation:

5 0
4 years ago
Solve: (3x^2-y)dx + (4y^3-x)dy =0 and find the solution passing through (1,1).
nordsb [41]

Step-by-step explanation:

The given equation is

(3x^{2}-y)dx+(4y^{3}-x)dy=0\\M(x,y)dx+N(x,y)dy=0

As a check for exactness we have

\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial (4y^{3}-x)}{\partial x} =-1\\\\\frac{\partial M}{\partial y}=\frac{\partial (3x^{3}-y)}{\partial y} =-1\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}=-1

Hence the given equation is an exact differential equation and thus the solution is given by

thus the solution is given by

u(x,y)=\int M(x,y)\partial x+\phi (y)\\\\u(x,y)=\int (3x^{2}-y)\partial x+\phi (y,c)\\\\u(x,y)=x^{3}-xy+\phi (y,c)\\\\

Similarly we have

u(x,y)=\int N(x,y)\partial y+\phi (x,c)\\\\u(x,y)=\int (4y^{3}-x)\partial y+\phi (x,c)\\\\u(x,y)=y^{4}-xy+\phi (x,c)\\\\

Comparing both the solutions we infer

\phi (x,c)=x^{3}+c

Hence the solution becomes

u(x,y)=x^{3}+y^{4}-xy=c

given boundary condition is that it passes through (1,1) hence

1^{3}+1^{4}-1=c\\\\\therefore c=1

thus solution is

u(x,y)=x^{3}+y^{4}-xy=1

4 0
3 years ago
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