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andre [41]
3 years ago
11

Your math teacher is planning a test for you. The test will have 30 questions. Some of

Mathematics
1 answer:
marishachu [46]3 years ago
5 0

Answer:

20 questions for 3 points

10 questions for 4 points

Step-by-step explanation:

20 × 3 = 60

10 × 4 = 40

60 + 40 = 100

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kicyunya [14]
Divdie by 12
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7 0
3 years ago
Someone help me solve this
NikAS [45]

First you multiply the two equations, equaling 3000000 kg and 60000 kg respectively. Once you have those numbers, you divide the Oakland mass by the Mapleville mass, so 3000000/60000. The answer is 50, to the tree is B, 50 times as big.

6 0
4 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
How to prove this???
swat32
\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3)  \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 


go to right side now

4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)

use x^3 - y^3 = (x-y)(x^2 + xy + y^2) and x^3 + y^3 = x^2 - xy + y^2

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow  4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad  \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)

so \sin^2 A + \cos^2 A = 1

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side
4 0
3 years ago
Help help math please please
wlad13 [49]

\underline{\underline{\large\bf{Solution:-}}}\\

The standard form of a line is given by -

\green{ \underline { \boxed{ \sf{Ax+By=C}}}}

where

  • A= integer
  • x = x-intercept
  • B= integer
  • y = y-intercept
  • C = constant

<h3><u>Given Equation: - </u></h3>

\begin{gathered}\\\implies\quad \sf y =  -\frac{5x}{4} -1 \\\end{gathered}

Arranging equation in standard form -

\begin{gathered}\\\implies\quad \sf y =  -\frac{5x}{4} - \frac{1}{1} \quad (1 = \frac{1}{1})\\\end{gathered}

\begin{gathered}\\\implies\quad \sf y =  -\frac{5x}{4} - \frac{1 \times 4}{1 \times 4} \\\end{gathered}

\begin{gathered}\\\implies\quad \sf y =  -\frac{5x}{4} - \frac{4}{4} \\\end{gathered}

\begin{gathered}\\\implies\quad \sf y =  \frac{-5x-4}{4}  \\\end{gathered}

\begin{gathered}\\\implies\quad \sf 4y =  -5x-4  \\\end{gathered}

\begin{gathered}\\\implies\quad \sf 4y +5x =-4  \\\end{gathered}

\begin{gathered}\\\implies\quad \sf 5x +4y=-4  \\\end{gathered}

Comparing with standard form of line -

  • A = 5
  • B = 4

Thus , 5 is our required answer.

8 0
3 years ago
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