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guapka [62]
4 years ago
9

HELP ASAP PLEASE! Evaluate the numerical expression. − 1/3 − (− 4/9 )

Mathematics
2 answers:
MissTica4 years ago
8 0
I use KCC (Keep Change Change) so -1/3 would be kept the same, the minus would be turned positive and the -4/9 would be changed to positive 4/9.
Then it will look like this: -1/3+(+4/9) = 1/9
Hope this helps.
serious [3.7K]4 years ago
7 0
I gotcha covered.

So, let's separate the problem to its individual components.

We have -1/3

and we're subtracting (-)

- 4/9

Now, we can see that we have two minus signs right next to each other.

If that occurs, we can change the two minus signs to one positive sign. It's just a rule in math.

So, we can rewrite the problem:

-1/3 - (- 4/9) can be rewritten to:

-1/3 + 4/9

But!!! We're not finished yet. The denominators (that means bottom part of a fraction) are not the same.

The denominators most of the time have to be same.

We have a 3 and a 9...

How do we make them the same? I gotcha covered again.

There's the criss-cross method, also known as cross multiplication, but I won't go over that because there's an easier way for this problem.

Here's the easier way:

3 * ? = 9

? = 3

So, 3 * 3 = 9.

We can multiply the - 1/3 by 3 to get the common denominator of 9.

There's a rule for this too... what we multiply by the denominator, we also MUST multiply that same number to the numerator (the top part of a fraction).

So, - 1/3 * (3/3) = - 3/9.

We can then add the fractions together since they have the same denominator.

Here it is:

- 3/9 + 4/9

Since - 3/9 is negative, we subtract that from 4/9...

Rewritten, we have:

4/9 - 3/9 = 1/9.

The denominators don't get changed unless we multiply or divide, so we only do the math for the numerators.

1/9 is the answer.

Good luck, and if you have more questions then please leave a comment!



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Step-by-step explanation:

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2sin5\theta=0\\\\sin5\theta=0\\\\5\theta=sin^{-1}(0)=n\pi;\ \ \ \ n=0,1,2,3,..\\\\\theta=\frac{n\pi}{5}

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aliina [53]

Answer:

Step-by-step explanation:

x+i*y≠0 ==> x≠0 and y≠0

x-i*y≠0 ==> x≠0 and y≠0

We must exclude (0,0) as solution.

\dfrac{1}{x+i*y} +\dfrac{2}{x-i*y} =1+i\\\\\dfrac{1*(x-i*y)}{(x+i*y)(x-i*y)} +\dfrac{2*(x+i*y)}{(x-i*y)(x+i*y)} =1+i\\\\\\\dfrac{x-i*y+2x+2i*y}{x^2+y^2} =1+i\\\\3x+i*y=(1+i)(x^2+y^2)\\\\3x+i*y=(x^2+y^2)+i(x^2+y^2+i*(x^2+y^2)\\\\\left\{\begin{array}{ccc}3x&=&x^2+y^2\\y&=&x^2+y^2\\\end {array}\right.\\\\\\\left\{\begin{array}{ccc}y&=&3x\\3x&=&x^2+9x^2\\\end {array}\right.\\\\\\\left\{\begin{array}{ccc}y&=&3x\\x(10x-3)&=&0\\\end {array}\right.\\\\

\left\{\begin{array}{ccc}x&=&0\\y&=&0\\\end {array}\right.\ (to\ exclude) \ or \ \left\{\begin{array}{ccc}x&=&\dfrac{3}{10}\\\\y&=&\dfrac{9}{10}\\\end {array}\right.\\\\Sol=\{(\dfrac{3}{10},\dfrac{9}{10})\}\\

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3 years ago
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notka56 [123]

Answer:

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Step-by-step explanation:

5 0
3 years ago
When ms. shreve solved an equation in class, she checked her solution and found that it did not make the equation true! Examine
Alex777 [14]
The first part of the second line, she left the -5 there. The correct work and solution should be this:
5(2x-1)-3x=5x+9
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4 years ago
In the diagram, m &lt; ACB = 65.<br> m &lt; ECA = <br><br> A. 180<br> B. 115<br> C. 65<br> D. 90
Westkost [7]
<span>m < ACB = 65.
m < ECA = 180 - 65 = 115

answer
</span>m < ECA  = 115

<span>B. 115</span>
3 0
3 years ago
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