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3 years ago

Help me please Hurry

Mathematics

1 answer:

Georgia [21]3 years ago

5 0

For all questions, use the concept of angles at a point (360°).

I also suggest and recommend that you specify the questions you need help with. It is best if you don't ask your homework here, because homework should be done by you yourself.

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Which equation shows the substitution method being used to solv the system of equations?

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Should Clayton buy this shower or not and why?

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3 years ago

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Are the following rational or irrational?

Scilla [17]

$\sqrt[3]{25}-\text{irrational}\\\\\sqrt[3]{64}=4\ because\ 4^3=64-\text{rattional}\\\\\sqrt[3]{125}=5\ because\ 5^3=125-\text{rattional}$

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3 years ago

Please guys help me out with this

Kisachek [45]

When you have a negative exponent, you move the variable with the negative exponent to the other side of the fraction to make the exponent positive.

For example:

$x^{-2}$ or $\frac{x^{-2}}{1} =\frac{1}{x^2}$

$\frac{1}{2y^{-3}} =\frac{y^3}{2}$ (you don't move the "2" because "2" has an exponent of 1, only "y" has the negative exponent)

$\frac{5^{-2}}{y}=\frac{1}{(5^2)y} =\frac{1}{25y}$

When you multiply a variable with an exponent by a variable with an exponent, you add the exponents together. (But you can only combine the exponents when the variables are the same)

$x^2(y^3)=x^2y^3$ (You can't combine them because they have different variables of x and y)

$x^3(x^5)=x^{(3+5)}=x^8$

$y^3(y^2)=y^{(3+2)}=y^5$

There's two ways you can do this. Either by first making all the exponents positive then do subtraction, or multiply them then make all the exponents positive. I'm doing the 2nd way.

$2w^7u^{-2}w^{-6}*9y^{-6}*2y^9u$ You can combine the numbers 2 x 9 x 2 = 36

$36w^7u^{-2}w^{-6}y^{-6}y^9u$ Now multiply the terms with the same variables

$36(w^{7+(-6)})(u^{(-2+1)})(y^{(-6+9)})$

$36(w^1)(u^{-1})(y^3)$ Now make all the exponents positive

$\frac{36wy^3}{u}$

If you were to do subtraction, here is what you need to know:

When you divide a variable with an exponent by a variable with an exponent, you subtract the exponents together. (Reminder: they have to be the same variable in order to combine the exponents)

For example:

$\frac{x^3}{x^2} =x^{(3-2)}=x^1$ or x

$\frac{y^2}{y^4} =y^{(2-4)}=y^{-2}=\frac{1}{y^2}$

7 0

3 years ago

All steps for: x/x-2 + x-1/x+1= -1

lorasvet [3.4K]

Answer:

$\large\boxed{x=0\ \vee\ x=1}$

Step-by-step explanation:

$Domain:\\\\x-2\neq0\ \wedge\ x+1\neq0\\\\x\neq2\ \wedge\ x\neq-1\\\\\boxed{D:\ x\in\mathbb{R}-\{-1,\ 2\}}\\\\=============================$

$\dfrac{x}{x-2}+\dfrac{x-1}{x+1}=-1\qquad\text{subtract}\ \dfrac{x-1}{x+1}\ \text{from both sides}\\\\\dfrac{x}{x-2}=-1-\dfrac{x-1}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-(x+1)}{x+1}+\dfrac{-(x-1)}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-(x+1)-(x-1)}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-x-1-x+1}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-2x}{x+1}\qquad\text{cross multiply}$

$x(x+1)=-2x(x-2)\qquad\text{use the distributive property}\\\\(x)(x)+(x)(1)=(-2x)(x)+(-2x)(-2)\\\\x^2+x=-2x^2+4x\qquad\text{add}\ 2x^2\ \text{to both sides}\\\\3x^2+x=4x\qquad\text{subtract 4x from both sides}\\\\3x^2-3x=0\qquad\text{distributive}\\\\3x(x-1)=0\iff 3x=0\ \vee\ x-1=0\\\\x=0\in D\ \vee\ x=1\in D$

4 0

3 years ago

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