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Romashka [77]
3 years ago
6

Think your a pro at riddles?

Mathematics
1 answer:
kondor19780726 [428]3 years ago
6 0

Answer:

nine

Step-by-step explanation:

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What is the value of x in the equation 2(x-3)+9=3(x+1)+x?
Sergio039 [100]

Answer:

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Divide using long division.<br><br> <img src="https://tex.z-dn.net/?f=%282x%5E%7B3%7D%20%2B5x%5E%7B2%7D%20%20-7x%2B6%29" id="Tex
babunello [35]

Answer:

The quotient = 2x - 1 and the remainder = 4x + 2

Step-by-step explanation:

\frac{2x^{3}+5x^{2}-7x+6}{x^{2}+3x-4}=2x+\frac{-x^{2}+x+6}{x^{2}+3x-4}

\frac{-x^{2}+x+6 }{x^{2}+3x-4}=-1+\frac{4x+2}{x^{2}+3x-4}

The quotient = 2x - 1 and the remainder = 4x + 2

3 0
3 years ago
That’s all i need to know. Thanks!
Svet_ta [14]

Answer:

c = 5

Step-by-step explanation:

a^2 + b^2 = c^2

3^2 + 4^2 = c^2

9 + 16 = 25

c^2 = 25

sqrt of 25 = 5

c= 5

6 0
3 years ago
Read 2 more answers
9.88+(6,7-x) = 2,6 please help​
ddd [48]
Hello

First you have to remove your parenthesis and since there is a “+” in front of the parenthesis, the expressions remain the same. So:
9.88+6.7-x = 2,6

Now, you have to add the numbers. So:
16.58-x = 2,6

Next, you have to move the constant (x) to the right and change its sign since it changes side. So:
-x = 2,6-16,58

Now, calculate the difference:
-x = 13,98

And lastly, change the signs on both sides of the equation since - + - equals +. So:
x = 13,98



Hope you found this helpful
Have a good day
8 0
4 years ago
Consider f and c below. f(x, y, z) = yzexzi + exzj + xyexzk,
Jet001 [13]

Answer:

f(x,y,z)=ye^{xz}+C  

Step-by-step explanation:

We can write the given expression as :

\vec f(x,y,z)=yze^{xz}\,\vec\imath+e^{xz}\,\vec\jmath+xye^{xz}\,\vec k

As given,   f = ∇f.

∇f = \dfrac{\partial f}{\partial x}i  + \dfrac{\partial f}{\partial y}j  +\dfrac{\partial f}{\partial z}k 

We can write the partial derivative with respect to x, y and z.

\dfrac{\partial f}{\partial x}=yze^{xz}       ___(Equation 1)

\dfrac{\partial f}{\partial y}=e^{xz}               ______(Equation 2)

\dfrac{\partial f}{\partial z}=xye^{xz}             ______(Equation 3)

Take equation 2 and integrate with respect to y,

\dfrac{\partial f}{\partial y}=e^{xz}

f(x,y,z)=ye^{xz}+a(x,z)           ----------Equation 4

Derivate both sides w.r.t x , we get :

\frac{d}{dx}(yze^{xz})=yze^{xz}+\dfrac{\partial a}{\partial x}

or

\dfrac{\partial a}{\partial x}=0

integrate

a(x,z)=b(z)

put in equation 4 ,

we get :

f(x,y,z)=ye^{xz}+b(z)

take derivative wrt z

\frac{d}{dz} (ye^{xz}+b(z))\impliesxye^{xz}=xye^{xz}+\frac{db}{dz}

we can take here:

\frac{db}{dz} = 0

integrate:

\int\ {\frac{db}{dz} } \, =\int0

b(z) = C

The function can be written as :

from equation 4 :

f(x,y,z)=ye^{xz}+C  

Where C is a constant.

4 0
3 years ago
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